Class 9 Mathematics Exercise-2.3 NCERT Solutions Olympiad Board Exam

Chapter 2

Polynomials

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

16 Questions

35–50 min Ideal time

Q1 Now at

📘 Concept & Theory Multiplication of Polynomials using Identities ›

In this exercise, we use algebraic identities to multiply polynomials quickly and accurately. Instead of multiplying every term manually, identities help us reduce calculation time and avoid mistakes.

Important Algebraic Identities Used

Identity	Formula	Use
Product of two binomials	\[\small (x+a)(x+b)=x^2+(a+b)x+ab \]	Used when first terms are same.
Difference of squares	\[\small (a+b)(a-b)=a^2-b^2 \]	Used when signs are opposite.

🗺️ Solution Roadmap Step-by-step Plan ›

Compare the expression with the identity.
Identify values of \(a\) and \(b\).
Substitute values carefully.
Simplify the middle term.

✏️ Solution Complete Solution ›

Step-by-step Solution · 8 steps

(i) \((x+4)(x+10)\)
\[\small \begin{aligned} (x+4)(x+10) \end{aligned} \]
Using the identity:
\[\small (x+a)(x+b)=x^2+(a+b)x+ab\]
Comparing,
\[\small a=4,\qquad b=10\]
Substitute these values into the identity:
\[\small \begin{aligned} (x+4)(x+10) &=x^2+(4+10)x+(4)(10) \\ &=x^2+14x+40 \end{aligned} \]

💡 Answer Final Answer ›

Product: \(x^2+14x+40\)

📘 Concept & Theory (ii) \((x+8)(x-10)\) ›

The variable term is same in both brackets. So we again use:

\[\small (x+a)(x+b)=x^2+(a+b)x+ab \]

Negative signs must be handled carefully.

🗺️ Solution Roadmap Step-by-step Plan ›

Take the second constant as negative
Add constants carefully while forming middle term
Multiply constants for last term

✏️ Solution Complete Solution ›

Step-by-step Solution · 4 steps

Comparing with
\[\small (x+a)(x+b)=x^2+(a+b)x+ab\]
we get \(a=8,\qquad b=-10\)
Now substitute:
\[\small \begin{aligned} (x+8)(x-10) &=x^2+\left(8+(-10)\right)x+(8)(-10) \\ &=x^2+(-2)x-80 \\ &=x^2-2x-80 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(x^2-2x-80\)

📘 Concept & Theory (iii) \((3x+4)(3x-5)\) ›

Here the first terms are \(3x\), not simply \(x\). We first factor out \(3\) from each bracket so that the identity can be applied properly.

🗺️ Solution Roadmap Step-by-step Plan ›

Rewrite each bracket by factoring \(3\).
Apply the identity.
Simplify fractions carefully.
Multiply by outside factor.

✏️ Solution Complete Solution ›

Step-by-step Solution · 13 steps

\[\small \begin{aligned} (3x+4)(3x-5) \end{aligned} \]
Rewrite each bracket:
\[\small \begin{aligned} (3x+4) &=3\left(x+\frac{4}{3}\right) \\ (3x-5) &=3\left(x-\frac{5}{3}\right) \end{aligned} \]
Therefore,
\[\small \begin{aligned} (3x+4)(3x-5) &=3\left(x+\frac{4}{3}\right)\times 3\left(x-\frac{5}{3}\right) \\ &=9\left(x+\frac{4}{3}\right)\left(x-\frac{5}{3}\right) \end{aligned} \]
Using identity:
\[\small (x+a)(x+b)=x^2+(a+b)x+ab \]
Here,
\[\small a=\frac{4}{3}, \qquad b=-\frac{5}{3} \]
Substitute:
\[\small \begin{aligned} &=9\left[ x^2+ \left( \frac{4}{3}-\frac{5}{3} \right)x+ \left( \frac{4}{3}\times -\frac{5}{3} \right) \right] \\ &=9\left[ x^2-\frac{x}{3}-\frac{20}{9} \right] \end{aligned} \]
Multiply \(9\) inside:
\[\small \begin{aligned} &=9(x^2)-9\left(\frac{x}{3}\right)-9\left(\frac{20}{9}\right) \\ &=9x^2-3x-20 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(9x^2-3x-20\)

📘 Concept & Theory (iv) \(\left(y^2+\frac{3}{2}\right) \left(y^2-\frac{3}{2}\right)\) ›

The two brackets differ only in sign. Hence we use the identity:

\[\small (a+b)(a-b)=a^2-b^2 \]

🗺️ Solution Roadmap Step-by-step Plan ›

Identify common term as \(a\).
Identify constant term as \(b\).
Square both terms separately.
Subtract carefully.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Comparing with
\[\small (a+b)(a-b)=a^2-b^2\]
we get
\[\small a=y^2,\qquad b=\frac{3}{2}\]
Now apply identity:
\[\small \begin{aligned} \left(y^2+\frac{3}{2}\right) \left(y^2-\frac{3}{2}\right) &= \left(y^2\right)^2- \left(\frac{3}{2}\right)^2 \\ &=y^4-\frac{9}{4} \end{aligned} \]

💡 Answer Final Answer ›

Final Answer : \( y^4-\frac{9}{4}\)

📘 Concept & Theory (v) \((3-2x)(3+2x)\) ›

This is a direct example of difference of squares identity.

\[\small (a-b)(a+b)=a^2-b^2 \]

🗺️ Solution Roadmap Step-by-step Plan ›

Identify \(a\) and \(b\).
Square each term separately.
Subtract second square from first.

✏️ Solution Complete Solution ›

Step-by-step Solution · 6 steps

Comparing with
\[\small (a-b)(a+b)=a^2-b^2\]
we get
\[\small a=3,\qquad b=2x\]
Therefore
\[\small \begin{aligned} (3-2x)(3+2x) &=3^2-(2x)^2 \\ &=9-4x^2 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(9-4x^2\)

🎯 Exam Significance Exam Significance ›

Helps in fast polynomial multiplication.
Very important for Board Examinations because identities are frequently used in long-answer questions.
Useful in competitive exams like Olympiads, NTSE, SSC, Railway and foundation level JEE preparation.
Builds algebraic manipulation skills required in higher mathematics.

↑ Top

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Q2 →

📘 Concept & Theory (i) 103 x 107 ›

Both numbers are close to \(100\). Therefore, rewrite them around the common base \(100\).

\[\small 103=100+3 \] \[\small 107=100+7 \]

Now use the identity:

\[\small (x+a)(x+b)=x^2+(a+b)x+ab \]

🗺️ Solution Roadmap Step-by-step Plan ›

Rewrite numbers using common base \(100\).
Identify \(x\), \(a\), and \(b\).
Apply identity carefully.
Simplify step by step.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} 103\times 107 &=(100+3)(100+7) \end{aligned} \]
Using identity:
\[\small (x+a)(x+b)=x^2+(a+b)x+ab\]
Comparing,
\[\small x=100,\qquad a=3,\qquad b=7\]
Substitute the values:
\[\small \begin{aligned} 103\times 107 &=100^2+(3+7)(100)+(3)(7) \\ &=10000+(10)(100)+21 \\ &=10000+1000+21 \\ &=11021 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(103\times 107=11021\)

📘 Concept & Theory (ii) 95 x 96 ›

Both numbers are slightly smaller than \(100\).

\[\small 95=100-5 \] \[\small 96=100-4 \]

Therefore, use:

\[\small (x+a)(x+b)=x^2+(a+b)x+ab \]

Here the values of \(a\) and \(b\) will be negative.

🗺️ Solution Roadmap Step-by-step Plan ›

Rewrite numbers using common base \(100\).
Identify \(x\), \(a\), and \(b\) (note the negative signs).
Apply identity carefully.
Simplify step by step.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} 95\times 96 &=(100-5)(100-4) \end{aligned} \]
Using identity:
\[\small (x+a)(x+b)=x^2+(a+b)x+ab\]
Comparing,
\[\small x=100,\qquad a=-5,\qquad b=-4\]
Substitute the values:
\[\small \begin{aligned} 95\times 96 &=100^2+(-5-4)(100)+(-5)(-4) \\ &=10000+(-9)(100)+20 \\ &=10000-900+20 \\ &=9120 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(95\times 96=9120\)

📘 Concept & Theory (iii) 104 x 96 ›

Here, one number is greater than \(100\) and the other is less than \(100\). We can still use the same identity by rewriting both numbers around \(100\).

\[\small 104=100+4 \] \[\small 96=100-4 \]

Now apply the identity:

\[\small (x+a)(x+b)=x^2+(a+b)x+ab \]

🗺️ Solution Roadmap Step-by-step Plan ›

Rewrite numbers using common base \(100\).
Identify \(x\), \(a\), and \(b\) (note the signs).
Apply identity carefully.
Simplify step by step.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} 104\times 96 &=(100+4)(100-4) \end{aligned} \]
Using identity:
\[\small (x+a)(x+b)=x^2+(a+b)x+ab\]
Comparing,
\[\small x=100,\qquad a=4,\qquad b=-4\]
Substitute the values:
\[\small \begin{aligned} 104\times 96 &=100^2+(4-4)(100)+(4)(-4) \\ &=10000+(0)(100)-16 \\ &=9984 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(104\times 96=9984\)

🎯 Exam Significance Exam Significance ›

Improves mental mathematics and algebraic thinking.
Frequently used in Olympiads, NTSE, SSC and aptitude tests.
Saves time in competitive entrance examinations.

← Q1

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Q3 →

📘 Concept & Theory (i) \(9x^2+6xy+y^2\) ›

We check whether the expression matches the identity:

\[\small a^2+2ab+b^2=(a+b)^2 \]

Here:

\(9x^2=(3x)^2\)
\(y^2=(y)^2\)
\(6xy=2(3x)(y)\)

Therefore, it is a perfect square trinomial.

🗺️ Solution Roadmap Step-by-step Plan ›

Express first and last terms as squares.
Check middle term carefully.
Match with identity.
Write factorised form.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

\[\small \begin{aligned} 9x^2+6xy+y^2 \end{aligned} \]
Rewrite the terms:
\[\small \begin{aligned} 9x^2&=(3x)^2 \\&=2(3x)(y) \\y^2 &=(y)^2 \end{aligned} \]
Therefore,
\[\small \begin{aligned} 9x^2+6xy+y^2 &=(3x)^2+2(3x)(y)+y^2 \end{aligned} \]
Using identity:
\[\small a^2+2ab+b^2=(a+b)^2 \]
We get,
\[\small a=3x,\qquad b=y\]
Substitute:
\[\small (3x+y)^2\]

💡 Answer Final Answer ›

Final Answer: \((3x+y)^2\)

📘 Concept & Theory (ii) \(4y^2-4y+1\) ›

This is also a perfect square trinomial. We check whether it matches the identity:

\[\small a^2-2ab+b^2=(a-b)^2 \]

Here:

\(4y^2=(2y)^2\)
\(1=(1)^2\)
\(-4y=-2(2y)(1)\)

Therefore, it can be factorised as a square of a binomial.

🗺️ Solution Roadmap Step-by-step Plan ›

Express first and last terms as squares.
Check middle term carefully.
Match with identity.
Write factorised form.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

\[\small \begin{aligned} 4y^2-4y+1 \end{aligned} \]
Rewrite the terms:
\[\small \begin{aligned} 4y^2&=(2y)^2 \\-4y &=-2(2y)(1) \\1 &=(1)^2 \end{aligned} \]
Therefore,
\[\small \begin{aligned} 4y^2-4y+1 &=(2y)^2-2(2y)(1)+1^2 \end{aligned} \]
Using identity:
\[\small a^2-2ab+b^2=(a-b)^2 \]
We get,
\[\small a=2y,\qquad b=1\]
Substitute:
\[\small (2y-1)^2\]

💡 Answer Final Answer ›

Final Answer: \((2y-1)^2\)

📘 Concept & Theory (iii) \(x^2-\frac{y^2}{100}\) ›

This is a difference of squares. We can use the identity:

\[\small a^2-b^2=(a+b)(a-b) \]

Here:

\(x^2=(x)^2\)
\(\frac{y^2}{100}=\left(\frac{y}{10}\right)^2\)

Therefore, it can be factorised as a product of two binomials.

🗺️ Solution Roadmap Step-by-step Plan ›

Express both terms as squares.
Match with difference of squares identity.
Identify \(a\) and \(b\).
Write factorised form.

✏️ Solution Complete Solution ›

Step-by-step Solution · 9 steps

\[\small \begin{aligned} x^2-\frac{y^2}{100} \end{aligned} \]
Rewrite the terms:
\[\small \begin{aligned} x^2&=(x)^2 \\&=\left(\frac{y}{10}\right)^2 \end{aligned} \]
Using identity:
\[\small a^2-b^2=(a+b)(a-b) \]
We get,
\[\small a=x,\qquad b=\frac{y}{10}\]
Substitute:
\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)

💡 Answer Final Answer ›

Final Answer: \(\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)\)

🎯 Exam Significance Exam Significance ›

Factorisation is one of the most important algebraic skills in Class 9 Mathematics.
Frequently asked in Board Examinations for short and long-answer questions.
Useful in solving equations, simplifying expressions and polynomial division.
Forms the foundation for quadratic equations and higher algebra in Classes 10, 11 and 12.

← Q2

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Q4 →

📘 Concept & Theory (i) \((x+2y+4z)^2\) ›

The expression contains three terms added together. Therefore we use:

\[\small (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \]

Here:

\[\small a=x,\qquad b=2y,\qquad c=4z \]

🗺️ Solution Roadmap Step-by-step Plan ›

Square each individual term carefully.
Multiply each pair of terms and then double the product.
Pay special attention to negative signs.
Fractional coefficients must also be handled carefully.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} (x+2y+4z)^2 \end{aligned} \]
Using identity:
\[\small (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \]
Substitute values:
\[\small \begin{aligned} (x+2y+4z)^2 &= x^2+ (2y)^2+ (4z)^2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x) \end{aligned} \]
Now simplify step by step:
\[\small \begin{aligned} &= x^2+ 4y^2+ 16z^2+ 4xy+ 16yz+ 8xz \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(x^2+4y^2+16z^2+4xy+16yz+8xz\)

📘 Concept & Theory (ii) \((2x-y+z)^2\) ›

This is also an expansion of a trinomial square. We use the same identity:

\[\small (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \]

Here:

\[\small a=2x,\qquad b=-y,\qquad c=z \]

🗺️ Solution Roadmap Step-by-step Plan ›

Square each individual term carefully.
Multiply each pair of terms and then double the product.
Pay special attention to negative signs.
Fractional coefficients must also be handled carefully.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} (2x-y+z)^2 \end{aligned} \]
Using identity:
\[\small (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \]
Substitute values:
\[\small \begin{aligned} (2x-y+z)^2 &= (2x)^2+ (-y)^2+ (z)^2 + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x) \end{aligned} \]
Now simplify step by step:
\[\small \begin{aligned} &= 4x^2+ y^2+ z^2- 4xy- 2yz+ 4xz \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(4x^2+y^2+z^2-4xy-2yz+4xz\)

📘 Concept & Theory (iii) \((–2x+3y+2z)^2\) ›

This is also an expansion of a trinomial square. We use the same identity:

\[\small (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \]

Here:

\[\small a=-2x,\qquad b=3y,\qquad c=2z \]

🗺️ Solution Roadmap Step-by-step Plan ›

Square each individual term carefully.
Multiply each pair of terms and then double the product.
Pay special attention to negative signs.
Fractional coefficients must also be handled carefully.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} (–2x+3y+2z)^2 \end{aligned} \]
Using identity:
\[\small (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \]
Substitute values:
\[\small \begin{aligned} (–2x+3y+2z)^2 &= (-2x)^2+ (3y)^2+ (2z)^2 + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x) \end{aligned} \]
Now simplify step by step:
\[\small \begin{aligned} &= 4x^2+ 9y^2+ 4z^2- 12xy+ 12yz- 8xz \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(4x^2+9y^2+4z^2-12xy+12yz-8xz\)

📘 Concept & Theory (iv) \((3a-7b-c)^2\) ›

This is another expansion of a trinomial square. We use the same identity:

\[\small (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \]

Here:

\[\small a=3a,\qquad b=-7b,\qquad c=-c \]

🗺️ Solution Roadmap Step-by-step Plan ›

Square each individual term carefully.
Multiply each pair of terms and then double the product.
Pay special attention to negative signs.
Fractional coefficients must also be handled carefully.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} (3a-7b-c)^2 \end{aligned} \]
Using identity:
\[\small (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \]
Substitute values:
\[\small \begin{aligned} (3a-7b-c)^2 &= (3a)^2+ (-7b)^2+ (-c)^2 + 2(3a)(-7b) + 2(-7b)(-c) + 2(-c)(3a) \end{aligned} \]
Now simplify step by step:
\[\small \begin{aligned} &= 9a^2+ 49b^2+ c^2- 42ab+ 14bc- 6ac \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(9a^2+49b^2+c^2-42ab+14bc-6ac\)

← Q3

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Q5 →

📘 Concept & Theory (i) \( 4x^2 + 9y^2 + 16z^2 + 12xy – 24yz – 16xz\) ›

We first identify the square terms:

\[\small 4x^2=(2x)^2 \] \[\small 9y^2=(3y)^2 \] \[\small 16z^2=(4z)^2 \]

Now check the middle terms:

\[\small 12xy=2(2x)(3y) \] \[\small -24yz=2(3y)(-4z) \] \[\small -16xz=2(2x)(-4z) \]

Hence the expression matches:

\[\small a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c)^2 \]

🗺️ Solution Roadmap Step-by-step Plan ›

Express all square terms properly.
Verify each middle term carefully.
Match with trinomial square identity.
Write final factorised form.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

\[\small \begin{aligned} 4x^2+9y^2+16z^2+12xy-24yz-16xz \end{aligned} \]
Rewrite the expression:
\[\small \begin{aligned} &= (2x)^2+ (3y)^2+ (4z)^2 \\ &\quad + 2(2x)(3y) + 2(3y)(4z) + 2(2x)(4z) \end{aligned} \]
Using identity:
\[\small a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c)^2 \]
where
\[\small a=2x,\qquad b=3y,\qquad c=4z \]
Therefore,
\[\small \begin{aligned} 4x^2+9y^2+16z^2+12xy-24yz-16xz &= (2x+3y-4z)^2 \end{aligned} \]
Writing completely in factorised form:
\[\small \begin{aligned} &= (2x+3y-4z)(2x+3y-4z) \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \((2x+3y-4z)(2x+3y-4z)\)

📘 Concept & Theory (ii) \(2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8xz\) ›

We first identify the square terms:

\[\small 2x^2=(\sqrt{2}x)^2 \] \[\small y^2=(y)^2 \] \[\small 8z^2=(2\sqrt{2}z)^2 \]

Now check the middle terms:

\[\small -2\sqrt{2}xy=2(\sqrt{2}(-x))(y) \] \[\small 4\sqrt{2}yz=2(-y)(2\sqrt{2}z) \] \[\small -8xz=2(\sqrt{2}(-x))(2\sqrt{2}z) \]

Hence the expression matches:

\[\small a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c)^2 \]

🗺️ Solution Roadmap Step-by-step Plan ›

Express all square terms properly.
Verify each middle term carefully.
Match with trinomial square identity.
Write final factorised form.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

\[\small \begin{aligned} 2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz \end{aligned} \]
Rewrite the expression:
\[\small \begin{aligned} &= (\sqrt{2}x)^2+ (y)^2+ (2\sqrt{2}z)^2 \\ &\quad + 2(\sqrt{2}(-x))(y) + 2(y)(2\sqrt{2}z) + 2(\sqrt{2}(-x))(2\sqrt{2}z) \end{aligned} \]
Using identity:
\[\small a^2+b^2+c^2+2ab+2bc+2ca=(a+b+c)^2 \]
where
\[\small a=\sqrt{2}(-x),\qquad b=y,\qquad c=2\sqrt{2}z \]
Therefore,
\[\small \begin{aligned} 2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz &= (\sqrt{2}(-x)+y+2\sqrt{2}z)^2 \end{aligned} \]
Writing completely in factorised form:
\[\small \begin{aligned} &= (\sqrt{2}(-x)-y+2\sqrt{2}z)(\sqrt{2}(-x)-y+2\sqrt{2}z) \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \((\sqrt{2}(-x)+y+2\sqrt{2}z)(\sqrt{2}(-x)+y+2\sqrt{2}z)\)

🎯 Exam Significance Exam Significance ›

Frequently asked in Board Examination algebra sections.
Helps in simplifying complicated polynomial expressions quickly.
Important for higher algebra, quadratic equations and coordinate geometry.
Improves pattern recognition skills required in Olympiads and aptitude tests.

← Q4

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Q6 →

📘 Concept & Theory (i) \( (2x + 1)3\) ›

The expression is in the form:

\[\small (a+b)^3 \] Therefore we use: \[\small (a+b)^3 = a^3+b^3+3ab(a+b) \] Here, \[\small a=2x, \qquad b=1 \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} (2x+1)^3 \end{aligned} \]
Using identity:
\[\small (a+b)^3=a^3+b^3+3ab(a+b)\]
Substitute values:
\[\small \begin{aligned} (2x+1)^3 &= (2x)^3+ (1)^3+ 3(2x)(1)((2x)+(1)) \end{aligned}\]
Now simplify step by step:
\[\small \begin{aligned} &= 8x^3+ 1+ 3(2x)(1)(2x+1) \\ &= 8x^3+ 1+ 6x(2x+1) \\ &= 8x^3+ 1+ 12x^2+ 6x \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(8x^3+12x^2+6x+1\)

📘 Concept & Theory (ii) \( (2a – 3b)3\) ›

The expression is in the form:

\[\small (a-b)^3 \] Therefore we use: \[\small (a-b)^3 = a^3-b^3-3ab(a-b) \] Here, \[\small a=2a, \qquad b=3b \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} (2a-3b)^3 \end{aligned} \]
Using identity:
\[\small (a-b)^3=a^3-b^3-3ab(a-b)\]
Substitute values:
\[\small \begin{aligned} (2a-3b)^3 &= (2a)^3- (3b)^3- 3(2a)(3b)((2a)-(3b)) \end{aligned}\]
Now simplify step by step:
\[\small \begin{aligned} &= 8a^3- 27b^3- 18ab(2a-3b) \\ &= 8a^3- 27b^3- 36a^2b+ 54ab^2 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(8a^3-27b^3-36a^2b+54ab^2\)

📘 Concept & Theory (iii) \(\left[\frac{3}{2}x+1\right]^3\) ›

The expression is in the form:

\[\small (a+b)^3 \] Therefore we use: \[\small (a+b)^3 = a^3+b^3+3ab(a+b) \] Here, \[\small a=\frac{3}{2}x, \qquad b=1 \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} \left[\frac{3}{2}x+1\right]^3 \end{aligned} \]
Using identity:
\[\small (a+b)^3=a^3+b^3+3ab(a+b)\]
Substitute values:
\[\small \begin{aligned} \left[\frac{3}{2}x+1\right]^3 &= \left(\frac{3}{2}x\right)^3+ (1)^3+ 3\left(\frac{3}{2}x\right)(1)\left(\frac{3}{2}x+1\right) \end{aligned}\]
Now simplify step by step:
\[\small \begin{aligned} &= \frac{27}{8}x^3+ 1+ \frac{9}{2}x\left(\frac{3}{2}x+1\right) \\ &= \frac{27}{8}x^3+ 1+ \frac{27}{4}x^2+ \frac{9}{2}x \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(\frac{27}{8}x^3+\frac{27}{4}x^2+\frac{9}{2}x+1\)

📘 Concept & Theory (iv) \(\left[x-\frac{2}{3}y\right]^3\) ›

Since subtraction is involved, we use:

\[\small (a-b)^3 = a^3-b^3-3ab(a-b) \] Here, \[\small a=x, \qquad b=\frac{2}{3}y \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} \left[x-\frac{2}{3}y\right]^3 \end{aligned} \]
Using identity:
\[\small (a-b)^3=a^3-b^3-3ab(a-b)\]
Substitute values:
\[\small \begin{aligned} \left[x-\frac{2}{3}y\right]^3 &= (x)^3- \left(\frac{2}{3}y\right)^3- 3(x)\left(\frac{2}{3}y\right)\left(x-\frac{2}{3}y\right) \end{aligned}\]
Now simplify step by step:
\[\small \begin{aligned} &= x^3- \frac{8}{27}y^3- 2xy\left(x-\frac{2}{3}y\right) \\ &= x^3- \frac{8}{27}y^3- 2x^2y+ \frac{4}{3}xy^2 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(x^3-\frac{8}{27}y^3-2x^2y+\frac{4}{3}xy^2\)

🎯 Exam Significance Exam Significance ›

Frequently asked in Board Examination algebra sections.
Useful in simplification and polynomial expansion problems.
Forms the foundation for higher algebra and binomial expansion.
Important for Olympiads and competitive aptitude exams.

← Q5

6 / 16 · 38%

Q7 →

📘 Concept & Theory (i) \( (99)^3\) ›

The number \(99\) is very close to \(100\).

\[\small 99=100-1 \]

Therefore we use:

\[\small (a-b)^3 = a^3-3a^2b+3ab^2-b^3 \]

Here:

\[\small a=100, \qquad b=1 \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} (99)^3 &= (100-1)^3 \end{aligned} \]
Using identity:
\[\small (a-b)^3 = a^3-3a^2b+3ab^2-b^3 \]
Substitute values:
\[\small \begin{aligned} (100-1)^3 &= 100^3 - 3(100)^2(1) + 3(100)(1)^2 - 1^3 \end{aligned} \]
Evaluate step by step:
\[\small \begin{aligned} &= 1000000 - 30000 + 300 - 1 \\ &= 970299 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(970299\)

📘 Concept & Theory (ii) \( (102)^3\) ›

The number \(102\) is very close to \(100\).

\[\small 102=100+2 \]

Therefore we use:

\[\small (a+b)^3 = a^3+3a^2b+3ab^2+b^3 \]

Here:

\[\small a=100, \qquad b=2 \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} (102)^3 &= (100+2)^3 \end{aligned} \]
Using identity:
\[\small (a+b)^3 = a^3+3a^2b+3ab^2+b^3 \]
Substitute values:
\[\small \begin{aligned} (100+2)^3 &= 100^3 + 3(100)^2(2) + 3(100)(2)^2 + 2^3 \end{aligned} \]
Evaluate step by step:
\[\small \begin{aligned} &= 1000000 + 60000 + 1200 + 8 \\ &= 1061208 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(1061208\)

📘 Concept & Theory (iii) \( (998)^3\) ›

The number \(998\) is close to \(1000\).

\[\small 998=1000-2 \]

Hence we use:

\[\small (a-b)^3 = a^3-3a^2b+3ab^2-b^3 \]

Here:

\[\small a=1000, \qquad b=2 \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} (998)^3 &= (1000-2)^3 \end{aligned} \]
Using identity:
\[\small (a-b)^3 = a^3-3a^2b+3ab^2-b^3 \]
Substitute values:
\[\small \begin{aligned} (1000-2)^3 &= 1000^3 - 3(1000)^2(2) + 3(1000)(2)^2 - 2^3 \end{aligned} \]
Evaluate step by step:
\[\small \begin{aligned} &= 1000000000 - 6000000 + 12000 - 8 \\ &= 994011992 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(994011992\)

🎯 Exam Significance Exam Significance ›

Improves fast calculation techniques.
Frequently used in Board Examination mental mathematics questions.
Helpful for aptitude tests and Olympiads.
Strengthens understanding of algebraic identities and expansions.

← Q6

7 / 16 · 44%

Q8 →

📘 Concept & Theory (i) \( 8a^3 + b^3 + 12a^2b + 6ab^2\) ›

We compare the expression with:

\[\small a^3+b^3+3a^2b+3ab^2=(a+b)^3 \]

Observe:

\[\small 8a^3=(2a)^3 \] \[\small 12a^2b=3(2a)^2(b) \] \[\small 6ab^2=3(2a)(b^2) \]

Hence the expression is a perfect cube.

✏️ Solution Complete Solution ›

Step-by-step Solution · 7 steps

\[\small \begin{aligned} 8a^3+b^3+12a^2b+6ab^2 \end{aligned} \]
Rewrite the terms:
\[\small \begin{aligned} &= (2a)^3+ b^3+ 3(2a)^2(b) + 3(2a)(b^2) \end{aligned} \]
Using identity:
\[\small a^3+b^3+3a^2b+3ab^2=(a+b)^3 \]
Therefore,
\[\small \begin{aligned} 8a^3+b^3+12a^2b+6ab^2 &= (2a+b)^3 \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \((2a+b)^3\)

📘 Concept & Theory (ii) \(8a^3 – b^3 – 12a^2b + 6ab^2\) ›

This expression matches:

\[\small a^3-b^3-3a^2b+3ab^2=(a-b)^3 \]

Observe:

\[\small 8a^3=(2a)^3 \] \[\small -12a^2b=-3(2a)^2(b) \] \[\small 6ab^2=3(2a)(b^2) \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 9 steps

\[\small \begin{aligned} 8a^3-b^3-12a^2b+6ab^2 \end{aligned} \]
Rewrite:
\[\small \begin{aligned} &= (2a)^3- b^3- 3(2a)^2(b) + 3(2a)(b^2) \end{aligned} \]
Using identity:
\[\small a^3-b^3-3a^2b+3ab^2=(a-b)^3\]
Therefore,
\[\small \begin{aligned} &= (2a-b)^3 \end{aligned} \]
Complete factorised form:
\[\small \begin{aligned} &= (2a-b)(2a-b)(2a-b) \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \((2a-b)(2a-b)(2a-b)\)

📘 Concept & Theory (iii) \(27 – 125a^3 – 135a + 225a^2\) ›

This expression matches:

\[\small a^3-b^3-3a^2b+3ab^2=(a-b)^3 \]

Observe:

\[\small 27=(3)^3 \] \[\small -125a^3=-(5a)^3 \] \[\small -135a=-3(5a)^2(3) \] \[\small 225a^2=3(5a)(3)^2 \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 9 steps

\[\small \begin{aligned} 27-125a^3-135a+225a^2 \end{aligned} \]
Rewrite:
\[\small \begin{aligned} &= (3)^3- (5a)^3- 3(5a)^2(3)+ 3(5a)(3)^2 \end{aligned} \]
Using identity:
\[\small a^3-b^3-3a^2b+3ab^2=(a-b)^3\]
Therefore,
\[\small \begin{aligned} &= (3-5a)^3 \end{aligned} \]
Complete factorised form:
\[\small \begin{aligned} &= (3-5a)(3-5a)(3-5a) \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \((3-5a)(3-5a)(3-5a)\)

📘 Concept & Theory (iv) \(64a^3 – 27b^3 – 144a^2b + 108ab^2\) ›

This expression matches:

\[\small a^3-b^3-3a^2b+3ab^2=(a-b)^3 \]

Observe:

\[\small 64a^3=(4a)^3 \] \[\small -27b^3=-(3b)^3 \] \[\small -144a^2b=-3(4a)^2(3b) \] \[\small 108ab^2=3(4a)(3b)^2 \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 9 steps

\[\small \begin{aligned} 64a^3-27b^3-144a^2b+108ab^2 \end{aligned} \]
Rewrite:
\[\small \begin{aligned} &= (4a)^3- (3b)^3- 3(4a)^2(3b)+ 3(4a)(3b)^2 \end{aligned} \]
Using identity:
\[\small a^3-b^3-3a^2b+3ab^2=(a-b)^3\]
Therefore,
\[\small \begin{aligned} &= (4a-3b)^3 \end{aligned} \]
Complete factorised form:
\[\small \begin{aligned} &= (4a-3b)(4a-3b)(4a-3b) \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \((4a-3b)(4a-3b)(4a-3b)\)

📘 Concept & Theory (v) \(27p^3-\frac{1}{216}-\frac{9}{2}p^2-\frac{1}{4}p\) ›

This expression matches:

\[\small a^3-b^3-3a^2b+3ab^2=(a-b)^3 \]

Observe:

\[\small 27p^3=(3p)^3 \] \[\small -\frac{1}{216}=-\left(\frac{1}{6}\right)^3 \] \[\small -\frac{9}{2}p^2=-3(3p)^2\left(\frac{1}{6}\right) \] \[\small -\frac{1}{4}p=3(3p)\left(\frac{1}{6}\right)^2 \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 9 steps

\[\small \begin{aligned} 27p^3-\frac{1}{216}-\frac{9}{2}p^2-\frac{1}{4}p \end{aligned} \]
Rewrite:
\[\small \begin{aligned} &= (3p)^3- \left(\frac{1}{6}\right)^3- 3(3p)^2\left(\frac{1}{6}\right)+ 3(3p)\left(\frac{1}{6}\right)^2 \end{aligned} \]
Using identity:
\[\small a^3-b^3-3a^2b+3ab^2=(a-b)^3\]
Therefore,
\[\small \begin{aligned} &= (3p-\frac{1}{6})^3 \end{aligned} \]
Complete factorised form:
\[\small \begin{aligned} &= (3p-\frac{1}{6})(3p-\frac{1}{6})(3p-\frac{1}{6}) \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \((3p-\frac{1}{6})(3p-\frac{1}{6})(3p-\frac{1}{6})\)

🎯 Exam Significance Exam Significance ›

Perfect cube recognition is an important algebraic skill.
Frequently asked in Board Examinations and scholarship tests.
Helps in simplification and higher polynomial factorisation.
Builds strong understanding of algebraic identities for Classes 10–12.

← Q7

8 / 16 · 50%

Q9 →

📘 Concept & Theory (i) \(x^3 + y^3 = (x + y) (x^2 – xy + y^2)\) ›

To verify the identity, we expand the RHS using distributive law.

If after simplification RHS becomes equal to LHS, the identity is verified.

🗺️ Solution Roadmap Step-by-step Plan ›

Expand the brackets carefully.
Multiply each term systematically.
Combine like terms.
Cancel opposite terms if possible.
Compare with LHS.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

h6{Expanding RHS:}
\[\small \begin{aligned} (x+y)(x^2-xy+y^2) \end{aligned} \]
Multiply \(x\) with each term:
\[\small \begin{aligned} &= x(x^2) + x(-xy) + x(y^2) \\ &\quad + y(x^2) + y(-xy) + y(y^2) \end{aligned} \]
Simplify each product:
\[\small \begin{aligned} &= x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3 \end{aligned} \]
Rearrange terms:
\[\small \begin{aligned} &= x^3+y^3 - x^2y+x^2y + xy^2-xy^2 \end{aligned} \]
Cancel opposite terms:
\[\small \begin{aligned} &= x^3+y^3+0+0 \end{aligned} \] Therefore, \[\small \begin{aligned} &= x^3+y^3 \end{aligned} \]
Since RHS = LHS

💡 Answer Final Answer ›

Hence Verified

📘 Concept & Theory (ii) \(x^3 – y^3 = (x – y) (x^2 + xy + y^2)\) ›

Similar to the previous part, we will expand the RHS and compare with LHS.

🗺️ Solution Roadmap Step-by-step Plan ›

Expand the brackets carefully.
Multiply each term systematically.
Combine like terms.
Cancel opposite terms if possible.
Compare with LHS.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

h6{Expanding RHS:}
\[\small \begin{aligned} (x-y)(x^2+xy+y^2) \end{aligned} \]
Multiply \(x\) with each term:
\[\small \begin{aligned} &= x(x^2) + x(xy) + x(y^2) \\ &\quad - y(x^2) - y(xy) - y(y^2) \end{aligned} \]
Simplify each product:
\[\small \begin{aligned} &= x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 \end{aligned} \]
Rearrange terms:
\[\small \begin{aligned} &= x^3-y^3 + x^2y-x^2y + xy^2-xy^2 \end{aligned} \]
Cancel opposite terms:
\[\small \begin{aligned} &= x^3-y^3+0+0 \end{aligned} \]
Since RHS = LHS

💡 Answer Final Answer ›

Hence Verified

🎯 Exam Significance Exam Significance ›

Verification problems improve algebraic manipulation skills.
Frequently asked in Board Examinations.
Helps in understanding polynomial factorisation deeply.
Important foundation for higher algebra and equation solving.

← Q8

9 / 16 · 56%

Q10 →

📘 Concept & Theory (i) \(27y^3 + 125z^3\) ›

First express each term as a perfect cube.

\[\small 27y^3=(3y)^3 \] \[\small 125z^3=(5z)^3 \]

Since both terms are added, we use:

\[\small a^3+b^3 = (a+b)(a^2-ab+b^2) \]

🗺️ Solution Roadmap Step-by-step Plan ›

Identify cube roots of each term.
Apply sum of cubes identity.
Substitute carefully.
Simplify the quadratic factor.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

\[\small \begin{aligned} 27y^3+125z^3 \end{aligned} \]
Express each term as cubes:
\[\small \begin{aligned} &= (3y)^3+ (5z)^3 \end{aligned} \]
Using identity:
\[\small a^3+b^3 = (a+b)(a^2-ab+b^2) \]
where
\[\small a=3y, \qquad b=5z \]
Therefore,
\[\small \begin{aligned} (3y)^3+(5z)^3 &= (3y+5z) \left[ (3y)^2-(3y)(5z)+(5z)^2 \right] \end{aligned} \]
Simplify each term:
\[\small \begin{aligned} &= (3y+5z) \left( 9y^2-15yz+25z^2 \right) \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \((3y+5z)(9y^2-15yz+25z^2)\)

📘 Concept & Theory (ii) \(64m^3 – 343n^3\) ›

First express each term as a perfect cube.

\[\small 64m^3=(4m)^3 \] \[\small 343n^3=(7n)^3 \]

Since the terms are subtracted, we use:

\[\small a^3-b^3 = (a-b)(a^2+ab+b^2) \]

🗺️ Solution Roadmap Step-by-step Plan ›

Identify cube roots of each term.
Apply difference of cubes identity.
Substitute carefully.
Simplify the quadratic factor.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

\[\small \begin{aligned} 64m^3-343n^3 \end{aligned} \]
Express each term as cubes:
\[\small \begin{aligned} &= (4m)^3- (7n)^3 \end{aligned} \]
Using identity:
\[\small a^3-b^3 = (a-b)(a^2+ab+b^2) \]
where
\[\small a=4m, \qquad b=7n \]
Therefore,
\[\small \begin{aligned} (4m)^3-(7n)^3 &= (4m-7n) \left[ (4m)^2+(4m)(7n)+(7n)^2 \right] \end{aligned} \]
Simplify each term:
\[\small \begin{aligned} &= (4m-7n) \left( 16m^2+28mn+49n^2 \right) \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \((4m-7n)(16m^2+28mn+49n^2)\)

🎯 Exam Significance Exam Significance ›

Sum and difference of cubes are very important algebraic identities.
Frequently asked in Board Examination factorisation questions.
Useful in polynomial simplification and equation solving.
Important foundation for higher algebra and calculus.

← Q9

10 / 16 · 63%

Q11 →

📘 Concept & Theory \(27x^3 + y^3 + z^3 – 9xyz\) ›

We first rewrite the expression in the standard form:

\[\small a^3+b^3+c^3-3abc \]

Observe carefully:

\[\small 27x^3=(3x)^3 \]

Therefore,

\[\small 27x^3+y^3+z^3-9xyz = (3x)^3+y^3+z^3-3(3x)(y)(z) \]

Now the identity can be applied directly.

🗺️ Solution Roadmap Step-by-step Plan ›

Rewrite the expression in standard identity form.
Identify \(a\), \(b\), and \(c\).
Apply the identity carefully.
Simplify the final factorised form.

✏️ Solution Complete Solution ›

Step-by-step Solution · 11 steps

\[\small \begin{aligned} 27x^3+y^3+z^3-9xyz \end{aligned} \]
Rewrite:
\[\small \begin{aligned} &= (3x)^3+y^3+z^3-3(3x)(y)(z) \end{aligned} \]
Comparing with the identity:
\[\small a^3+b^3+c^3-3abc \]
we identify:
\[\small a=3x, \qquad b=y, \qquad c=z \]
Using the factorisation identity:
\[\small a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \]
Therefore,
\[\small \begin{aligned} 27x^3+y^3+z^3-9xyz &= \frac{1}{2} (3x+y+z) \\ &\quad \times \left[ (3x-y)^2+ (y-z)^2+ (z-3x)^2 \right] \end{aligned} \]

💡 Answer Final Answer ›

Final Answer: \(\frac{1}{2}(3x+y+z)\left[(3x-y)^2+(y-z)^2+(z-3x)^2\right]\)

🎯 Exam Significance Exam Significance ›

It helps factorise cubic expressions containing three variables.
It is frequently used in Olympiads and algebraic proofs.
It plays an important role in inequalities and polynomial theory.
It strengthens understanding of symmetric algebraic expressions.

← Q10

11 / 16 · 69%

Q12 →

📘 Concept & Theory To verify the identity, we simplify the RHS completely and show that it becomes equal to the LHS. ›

Expand each squared term carefully.
Combine like terms.
Multiply the resulting expression by \((x+y+z)\).
Cancel opposite terms systematically.

✏️ Solution Complete Solution ›

Step-by-step Solution · 21 steps

Verification:
RHS
\[\small \begin{aligned} &= \frac{1}{2} (x+y+z) \left[ (x-y)^2+ (y-z)^2+ (z-x)^2 \right] \end{aligned} \]
Expand each square:
\[\small \begin{aligned} &= \frac{1}{2} (x+y+z) \Big[ x^2+y^2-2xy \\ &\qquad + y^2+z^2-2yz \\ &\qquad + z^2+x^2-2zx \Big] \end{aligned} \]
Combine like terms:
\[\small \begin{aligned} &= \frac{1}{2} (x+y+z) \Big[ 2x^2+2y^2+2z^2 \\ &\qquad -2xy-2yz-2zx \Big] \end{aligned} \]
Take common factor \(2\):
\[\small \begin{aligned} &= \frac{1}{2} (x+y+z) \cdot 2 \left[ x^2+y^2+z^2-xy-yz-zx \right] \end{aligned} \]
Cancel \(\frac12\) and \(2\):
\[\small \begin{aligned} &= (x+y+z) (x^2+y^2+z^2-xy-yz-zx) \end{aligned} \]
Multiply term-by-term:
\[\small \begin{aligned} &= x(x^2+y^2+z^2-xy-yz-zx) \\ &\quad + y(x^2+y^2+z^2-xy-yz-zx) \\ &\quad + z(x^2+y^2+z^2-xy-yz-zx) \end{aligned} \]
Expand completely:
\[\small \begin{aligned} &= x^3+xy^2+xz^2-x^2y-xyz-x^2z \\ &\quad + x^2y+y^3+yz^2-xy^2-y^2z-xyz \\ &\quad + x^2z+y^2z+z^3-xyz-yz^2-z^2x \end{aligned} \]
Rearrange like terms:
\[\small \begin{aligned} &= x^3+y^3+z^3 \\ &\quad + (xy^2-xy^2) \\ &\quad + (xz^2-z^2x) \\ &\quad + (x^2y-x^2y) \\ &\quad + (x^2z-x^2z) \\ &\quad + (yz^2-yz^2) \\ &\quad + (y^2z-y^2z) \\ &\quad -xyz-xyz-xyz \end{aligned} \]
Cancel opposite terms:
\[\small \begin{aligned} &= x^3+y^3+z^3-3xyz \end{aligned} \]
Since
\[\small \text{RHS}=\text{LHS} \]

💡 Answer Final Answer ›

Hence Verified

← Q11

12 / 16 · 75%

Q13 →

📘 Concept & Theory Applications of the Identity ›

We use the identity:

\[\small x^3+y^3+z^3-3xyz = \frac{1}{2} (x+y+z) \left[ (x-y)^2+ (y-z)^2+ (z-x)^2 \right] \]

Since \(x+y+z=0\), the entire RHS becomes zero.

🗺️ Solution Roadmap Step-by-step Plan ›

Write the standard identity.
Substitute the given condition.
Simplify RHS.
Rearrange terms to obtain required result.

✏️ Solution Complete Solution ›

Step-by-step Solution · 10 steps

Using identity:
\[\small x^3+y^3+z^3-3xyz = \frac{1}{2} (x+y+z) \left[ (x-y)^2+ (y-z)^2+ (z-x)^2 \right] \]
Given:
\[\small x+y+z=0 \]
Substituting in the identity:
\[\small x^3+y^3+z^3-3xyz = \frac{1}{2} (0) \left[ (x-y)^2+ (y-z)^2+ (z-x)^2 \right] \]
Simplifying:
\[\small x^3+y^3+z^3-3xyz = 0 \]
Rearranging:
\[\small x^3+y^3+z^3 = 3xyz \]

💡 Answer Final Answer ›

Hence Proved

🎯 Exam Significance Exam Significance ›

Frequently used in algebraic proofs and simplifications.
Important for Olympiads and scholarship examinations.
Saves time in mental mathematics problems.
Useful in higher algebra and polynomial theory.

← Q12

13 / 16 · 81%

Q14 →

📘 Concept & Theory \( (–12)^3 + (7)^3 + (5)^3\) ›

We first check whether the sum of the three numbers is zero.

\[\small -12+7+5=0 \]

Therefore, we can use:

\[\small x^3+y^3+z^3=3xyz \] whenever \[\small x+y+z=0 \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 4 steps

Let \(x=-12\), \(y=7\), and \(z=5\).
Then, \(x+y+z=-12+7+5=0\).
Using the identity \(x^3+y^3+z^3=3xyz\):
\[\small (-12)^3 + (7)^3 + (5)^3 = 3(-12)(7)(5) = -1260 \]

💡 Answer Final Answer ›

Final Answer: \(-1260\)

📘 Concept & Theory \((28)^3 + (–15)^3 + (–13)^3\) ›

Similar to the previous part, we check if the sum is zero.

\[\small 28-15-13=0 \]

Therefore, we can apply the same identity.

✏️ Solution Complete Solution ›

Step-by-step Solution · 4 steps

Let \(x=28\), \(y=-15\), and \(z=-13\).
Then, \(x+y+z=28-15-13=0\).
Using the identity \(x^3+y^3+z^3=3xyz\):
\[\small (28)^3 + (-15)^3 + (-13)^3 = 3(28)(-15)(-13) = 16380 \]

💡 Answer Final Answer ›

Final Answer: \(16380\)

← Q13

14 / 16 · 88%

Q15 →

📘 Concept & Theory (i) \(\text{Area : }25a^2 – 35a + 12\) ›

We factorise the quadratic expression using the middle term splitting method.

For:

\[\small 25a^2-35a+12 \] we need two numbers whose:

Product \(=25\times12=300\)
Sum \(=-35\)

The required numbers are:

\[\small -15,\qquad -20 \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 18 steps

Given:
\[\small A=25a^2-35a+12\]
Find two numbers:
\[\small pq=25\times12=300\]
and
\[\small p+q=-35\]
The numbers are:
\[\small p=-15,\qquad q=-20\]
Split the middle term:
\[\small \begin{aligned} A &= 25a^2-15a-20a+12 \end{aligned} \]
Group terms:
\[\small \begin{aligned} &= 5a(5a-3)-4(5a-3) \end{aligned} \]
Take common factor:
\[\small \begin{aligned} &=(5a-3)(5a-4) \end{aligned} \]
Therefore,
\[\small \boxed{ \text{Length}=5a-3,\qquad \text{Breadth}=5a-4 } \]
Possible Dimensions:
\[\small (5a-3)\text{ units} \quad\text{and}\quad (5a-4)\text{ units}\]

💡 Answer Final Answer ›

Final Answer: Length = \(5a-3\) units, Breadth = \(5a-4\) units

📘 Concept & Theory (ii) \(\text{Area : }35y^2 + 13y –12\) ›

Similar to the previous part, we factorise the quadratic expression.

For:

\[\small 35y^2+13y-12 \] we need two numbers whose:

Product \(=35\times(-12)=-420\)
Sum \(=13\)

The required numbers are:

\[\small 28,\qquad -15 \]

✏️ Solution Complete Solution ›

Step-by-step Solution · 18 steps

Given:
\[\small A=35y^2+13y-12\]
Find two numbers:
\[\small pq=35\times(-12)=-420\]
and
\[\small p+q=13\]
The numbers are:
\[\small p=28,\qquad q=-15\]
Split the middle term:
\[\small \begin{aligned} A &= 35y^2+28y-15y-12 \end{aligned} \]
Group terms:
\[\small \begin{aligned} &= 7y(5y+4)-3(5y+4) \end{aligned} \]
Take common factor:
\[\small \begin{aligned} &=(5y+4)(7y-3) \end{aligned} \]
Therefore,
\[\small \boxed{ \text{Length}=5y+4,\qquad \text{Breadth}=7y-3 } \]
Possible Dimensions:
\[\small (5y+4)\text{ units} \quad\text{and}\quad (7y-3)\text{ units}\]

💡 Answer Final Answer ›

Final Answer: Length = \(5y+4\) units, Breadth = \(7y-3\) units

🎯 Exam Significance Exam Significance ›

Important application of factorisation in geometry.
Frequently asked in Board Examinations.
Strengthens algebraic reasoning and pattern recognition.
Useful foundation for quadratic equations and coordinate geometry.

← Q14

15 / 16 · 94%

Q16 →

📘 Concept & Theory (i) \(\text{Volume : }3x^2 – 12x\) ›

We factorise the expression by taking the common factor.

Both terms contain:

\[\small 3x \]

After factorisation, each factor represents one dimension of the cuboid.

✏️ Solution Complete Solution ›

Step-by-step Solution · 9 steps

Given:
\[\small V=3x^2-12x\]
Take common factor \(3x\):
\[\small V=3x(x-4)\]
Therefore, possible dimensions of the cuboid are:
\[\small \boxed{3x,\quad (x-4)}\]
Since volume of a cuboid has three dimensions, the third dimension may be taken as \(1\).
Hence possible dimensions are:
\[\small \boxed{3x,\quad (x-4),\quad 1}\]

💡 Answer Final Answer ›

Possible Dimensions: \(3x,\quad (x-4),\quad 1\)

📘 Concept & Theory (ii) \(\text{Volume : }12ky^2 + 8ky – 20k\) ›

First take out the common factor \(4k\).

Then factorise the quadratic expression:

\[\small 3y^2+2y-5 \] using middle-term splitting.

✏️ Solution Complete Solution ›

Step-by-step Solution · 22 steps

Given:
\[\small V=12ky^2+8ky-20k\]
Take common factor \(4k\):
\[\small V=4k(3y^2+2y-5)\]
Now factorise:
\[\small 3y^2+2y-5\]
Find two numbers whose:
\[\small \text{Product}=3\times(-5)=-15\]
and
\[\small \text{Sum}=2\]
The numbers are:
\[\small 5,\qquad -3\]
Split the middle term:
\[\small \begin{aligned} 3y^2+2y-5 &=3y^2+5y-3y-5 \end{aligned} \]
Group terms:
\[\small \begin{aligned} &=y(3y+5)-1(3y+5) \end{aligned} \]
Take common factor:
\[\small \begin{aligned} &=(y-1)(3y+5) \end{aligned} \]
Therefore,
\[\small \begin{aligned} V&=4k(y-1)(3y+5) \end{aligned} \]
Hence possible dimensions are:
\[\small \boxed{4k,\quad (y-1),\quad (3y+5)} \]

💡 Answer Final Answer ›

Possible Dimensions: \(4k,\quad (y-1),\quad (3y+5)\)

🎯 Exam Significance Exam Significance ›

Combines algebra with geometry applications.
Important for Board Examination word problems.
Strengthens factorisation skills.
Useful foundation for mensuration and coordinate geometry.

← Q15

16 / 16 · 100%

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