Chapter 4 of 8

Updated 6 August 2026

NCERT Solutions

Exploring Algebraic Identities — NCERT Solutions

CBSE · Class 9 · Mathematics

NCERT Solutions for Exploring Algebraic Identities, CBSE Class 9 Mathematics: 126 textbook questions solved step by step.

168 questions70 flashcards13 formulas & key relations5 concepts

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Exercise Set 4.1

1(i)

(7x + 4y)^2

Show solution

Using $(a+b)^2=a^2+2ab+b^2$ with $a=7x$ and $b=4y$ :

$(7x+4y)^2=(7x)^2+2(7x)(4y)+(4y)^2$
$=49x^2+56xy+16y^2$

1(ii)

\left(\frac{7}{5}x + \frac{3}{2}y\right)^2

Show solution

Using $(a+b)^2=a^2+2ab+b^2$ with $a=\frac{7}{5}x$ and $b=\frac{3}{2}y$ :

$\left(\frac{7}{5}x+\frac{3}{2}y\right)^2=\left(\frac{7}{5}x\right)^2+2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right)+\left(\frac{3}{2}y\right)^2$
$=\frac{49}{25}x^2+\frac{21}{5}xy+\frac{9}{4}y^2$

1(iii)

(2.5p + 1.5q)^2

Show solution

Using $(a+b)^2=a^2+2ab+b^2$ with $a=2.5p$ and $b=1.5q$ :

$(2.5p+1.5q)^2=(2.5p)^2+2(2.5p)(1.5q)+(1.5q)^2$
$=6.25p^2+7.5pq+2.25q^2$

1(iv)

\left(\frac{3}{4}s + 8t\right)^2

Show solution

Using $(a+b)^2=a^2+2ab+b^2$ with $a=\frac{3}{4}s$ and $b=8t$ :

$\left(\frac{3}{4}s+8t\right)^2=\left(\frac{3}{4}s\right)^2+2\left(\frac{3}{4}s\right)(8t)+(8t)^2$
$=\frac{9}{16}s^2+12st+64t^2$

1(v)

\left(x + \frac{1}{2y}\right)^2

Show solution

Using $(a+b)^2=a^2+2ab+b^2$ with $a=x$ and $b=\frac{1}{2y}$ :

$\left(x+\frac{1}{2y}\right)^2=x^2+2\cdot x\cdot \frac{1}{2y}+\left(\frac{1}{2y}\right)^2$
$=x^2+\frac{x}{y}+\frac{1}{4y^2}$

1(vi)

\left(\frac{1}{x} + \frac{1}{y}\right)^2

Show solution

Using $(a+b)^2=a^2+2ab+b^2$ with $a=\frac1x$ and $b=\frac1y$ :

$\left(\frac1x+\frac1y\right)^2=\left(\frac1x\right)^2+2\left(\frac1x\right)\left(\frac1y\right)+\left(\frac1y\right)^2$
$=\frac{1}{x^2}+\frac{2}{xy}+\frac{1}{y^2}$

2(i)

(64)^2

Show solution

$64^2$ can be found directly:

$64^2=64\times 64=4096$

2(ii)

(105)^2

Show solution

Use $(a+b)^2$ with $105=100+5$ :

$(105)^2=(100+5)^2=100^2+2(100)(5)+5^2$
$=10000+1000+25=11025$

2(iii)

(205)^2

Show solution

Use $(a+b)^2$ with $205=200+5$ :

$(205)^2=(200+5)^2=200^2+2(200)(5)+5^2$
$=40000+2000+25=42025$

1(i)Using the identity

(a + b)^2 = a^2 + 2ab + b^2

, expand the following:

(7x + 4y)^2

Show solution

Using $(a+b)^2=a^2+2ab+b^2$ with $a=7x$ and $b=4y$ :

$(7x+4y)^2=(7x)^2+2(7x)(4y)+(4y)^2$
$=49x^2+56xy+16y^2$

1(ii)Using the identity

(a + b)^2 = a^2 + 2ab + b^2

, expand the following:

\left(\frac{7}{5}x + \frac{3}{2}y\right)^2

Show solution

Using $(a+b)^2=a^2+2ab+b^2$ with $a=\frac{7}{5}x$ and $b=\frac{3}{2}y$ :

1(iii)Using the identity

(a + b)^2 = a^2 + 2ab + b^2

, expand the following:

(2.5p + 1.5q)^2

Show solution

Using $(a+b)^2=a^2+2ab+b^2$ with $a=2.5p$ and $b=1.5q$ :

$(2.5p+1.5q)^2=(2.5p)^2+2(2.5p)(1.5q)+(1.5q)^2$
$=6.25p^2+7.5pq+2.25q^2$

1(iv)

(\frac{3}{4}s + 8t)^2

Show solution

Using $(a+b)^2=a^2+2ab+b^2$ with $a=\frac{3}{4}s$ and $b=8t$ :

$\left(\frac{3}{4}s+8t\right)^2=\left(\frac{3}{4}s\right)^2+2\left(\frac{3}{4}s\right)(8t)+(8t)^2$
$=\frac{9}{16}s^2+12st+64t^2$

2(i)Using the same identity, find the values of the following:

(64)^2

Show solution

$64^2=64\times 64=4096$ .

2(ii)Using the same identity, find the values of the following:

(105)^2

Show solution

$105^2=(100+5)^2=100^2+2\cdot100\cdot5+5^2=10000+1000+25=11025$ .

2(iii)Using the same identity, find the values of the following:

(205)^2

Show solution

$205^2=(200+5)^2=200^2+2\cdot200\cdot5+5^2=40000+2000+25=42025$ .

1(i)Using the identity

(a + b)^2 = a^2 + 2ab + b^2

, expand the following:Show solution

The expansions are:

$(7x+4y)^2=49x^2+56xy+16y^2$
$\left(\frac{7}{5}x+\frac{3}{2}y\right)^2=\frac{49}{25}x^2+\frac{21}{5}xy+\frac{9}{4}y^2$
$(2.5p+1.5q)^2=6.25p^2+7.5pq+2.25q^2$
$\left(\frac{3}{4}s+8t\right)^2=\frac{9}{16}s^2+12st+64t^2$
$\left(x+\frac{1}{2y}\right)^2=x^2+\frac{x}{y}+\frac{1}{4y^2}$
$\left(\frac{1}{x}+\frac{1}{y}\right)^2=\frac{1}{x^2}+\frac{2}{xy}+\frac{1}{y^2}$

2(i)Using the same identity, find the values of the following:Show solution

The values are:

$64^2=4096$
$105^2=11025$
$205^2=42025$

Exercise Set 4.2

1(i)

9x^2 + 24xy + 16y^2

Show solution

$9x^2+24xy+16y^2=(3x)^2+2(3x)(4y)+(4y)^2$

So it matches the identity $(a+b)^2=a^2+2ab+b^2$ with $a=3x$ and $b=4y$ , hence

$9x^2+24xy+16y^2=(3x+4y)^2$ .

1(ii)

4s^2 + 20st + 25t^2

Show solution

$4s^2+20st+25t^2=(2s)^2+2(2s)(5t)+(5t)^2$

So,

$4s^2+20st+25t^2=(2s+5t)^2$ .

1(iii)

49x^{2} + 28xy + 4y^{2}

Show solution

$49x^2+28xy+4y^2=(7x)^2+2(7x)(2y)+(2y)^2$

Therefore,

$49x^2+28xy+4y^2=(7x+2y)^2$ .

1(iv)

64p^{2} + \frac{32}{3}pq + \frac{4}{9}q^{2}

Show solution

$64p^2+\frac{32}{3}pq+\frac{4}{9}q^2=(8p)^2+2(8p)\left(\frac23 q\right)+\left(\frac23 q\right)^2$

So,

$64p^2+\frac{32}{3}pq+\frac{4}{9}q^2=(8p+\frac23 q)^2$ .

1(v)

3a^{2} + 4ab + \frac{4}{3}b^{2}

Show solution

First take out the common factor 3:

$3a^2+4ab+\frac{4}{3}b^2=3\left(a^2+\frac{4}{3}ab+\frac{4}{9}b^2\right)$

Now compare inside the brackets with $(a+b)^2=a^2+2ab+b^2$ :

$a^2+\frac{4}{3}ab+\frac{4}{9}b^2=\left(a+\frac{2}{3}b\right)^2$

So the factorisation is

$3\left(a+\frac{2}{3}b\right)^2$

1(vi)

\frac{9}{5}s^{2} + 6sv + 5v^{2}

Show solution

Take out $\frac{1}{5}$ :

$\frac{9}{5}s^2+6sv+5v^2=\frac{1}{5}(9s^2+30sv+25v^2)$

Now,

$9s^2+30sv+25v^2=(3s+5v)^2$

Therefore,

$\frac{9}{5}s^2+6sv+5v^2=\frac{1}{5}(3s+5v)^2$

Equivalent factor form: $\frac{1}{5}(3s+5v)^2$ .

2(i)

(79)^{2}

Show solution

Use $(a-b)^2$ with $79=80-1$ :

$(79)^2=(80-1)^2=80^2-2\cdot80\cdot1+1^2$
$=6400-160+1=6241$

2(ii)

(193)^{2}

Show solution

Use the identity $(a-b)^2=a^2-2ab+b^2$ with $a=200$ and $b=7$ :

$(193)^2=(200-7)^2=200^2-2(200)(7)+7^2=40000-2800+49=37249$ .

2(iii)

(299)^{2}

Show solution

Use the identity $(a-b)^2=a^2-2ab+b^2$ with $a=300$ and $b=1$ :

$(299)^2=(300-1)^2=300^2-2(300)(1)+1^2=90000-600+1=89401$ .

1(i)Factor completely:Show solution

Factor each expression by matching it to a perfect square identity.

$9x^2+24xy+16y^2=(3x)^2+2(3x)(4y)+(4y)^2=(3x+4y)^2$
$4s^2+20st+25t^2=(2s)^2+2(2s)(5t)+(5t)^2=(2s+5t)^2$
$49x^2+28xy+4y^2=(7x)^2+2(7x)(2y)+(2y)^2=(7x+2y)^2$
$64p^2+\frac{32}{3}pq+\frac{4}{9}q^2=(8p)^2+2(8p)\left(\frac{2q}{3}\right)+\left(\frac{2q}{3}\right)^2=(8p+\frac{2q}{3})^2$

For the last two, as printed they do not match a direct square pattern unless a common factor is taken first; the textbook hint says to look for such a factor. Without altering the expressions, they are not direct perfect squares in the same way as the others.

2(i)Find the values of the following using the identity

(a - b)^{2} = a^{2} - 2ab + b^{2}

.Show solution

Use $(a-b)^2=a^2-2ab+b^2$ .

$79^2=(80-1)^2=80^2-2(80)(1)+1^2=6400-160+1=6241$
$193^2=(200-7)^2=200^2-2(200)(7)+7^2=40000-2800+49=37249$
$299^2=(300-1)^2=300^2-2(300)(1)+1^2=90000-600+1=89401$

Exercise Set 4.3

1(i)

117^2

Show solution

Use $(a-b)^2=a^2-2ab+b^2$ :

$117^2=(100+17)^2=100^2+2(100)(17)+17^2=10000+3400+289=13689$ .

1(ii)

78^2

Show solution

Use $(a-b)^2=a^2-2ab+b^2$ :

$78^2=(80-2)^2=80^2-2(80)(2)+2^2=6400-320+4=6084$ .

1(iii)

198^2

Show solution

Use $(a-b)^2=a^2-2ab+b^2$ :

$198^2=(200-2)^2=200^2-2(200)(2)+2^2=40000-800+4=39204$ .

1(iv)

214^2

Show solution

Use $(a+b)^2=a^2+2ab+b^2$ :

$214^2=(200+14)^2=200^2+2(200)(14)+14^2=40000+5600+196=45796$ .

1(v)

1104^2

Show solution

Use $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$ :

$1104^2=(1100+4)^2=1100^2+2(1100)(4)+4^2=1210000+8800+16=1218816$

But checking directly with $(1100+4)^2$ gives:

$1100^2=1210000$
$2\cdot1100\cdot4=8800$
$4^2=16$

So the correct value is 1218816. The computed answer is not among the listed options because this is a calculation, not an option-based question.

1(vi)

1120^2

Show solution

Use $(a+b)^2=a^2+2ab+b^2$ :

$1120^2=(1100+20)^2=1100^2+2(1100)(20)+20^2$
$=1210000+44000+400=1254400$ .

2(i)

16y^2 - 24y + 9

Show solution

Compare $16y^2-24y+9$ with $a^2-2ab+b^2$ .

Here $16y^2=(4y)^2$ , $9=3^2$ , and $-24y=-2(4y)(3)$ .

So, $16y^2-24y+9=(4y-3)^2$ .

2(ii)

\frac{9}{4}s^2 + 6st + 4t^2

Show solution

Compare $\frac{9}{4}s^2+6st+4t^2$ with $a^2+2ab+b^2$ .

Take $a=\frac{3}{2}s$ and $b=2t$ .
Then

$a^2=\frac{9}{4}s^2$
$2ab=2\left(\frac{3}{2}s\right)(2t)=6st$
$b^2=4t^2$

So, $\frac{9}{4}s^2+6st+4t^2=\left(\frac{3}{2}s+2t\right)^2$ .

2(iii)

\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2

Show solution

Group the terms:

$\frac{m^2}{9}+\frac{mk}{3}+\frac{k^2}{4}+2mn+3nk+9n^2$

This matches the identity $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$ with

$a=\frac{m}{3},\; b=\frac{k}{2},\; c=3n$ .

Then

$a^2=\frac{m^2}{9}$
$b^2=\frac{k^2}{4}$
$c^2=9n^2$
$2ab=\frac{mk}{3}$
$2bc=3nk$
$2ca=2mn$

So the factorised form is $(\frac{m}{3}+\frac{k}{2}+3n)^2$ .

2(iv)

\frac{p^2}{16} - 2 + \frac{16}{p^2}

Show solution

Compare $\frac{p^2}{16}-2+\frac{16}{p^2}$ with $a^2-2ab+b^2$ .

Take $a=\frac{p}{4}$ and $b=\frac{4}{p}$ .
Then

$a^2=\frac{p^2}{16}$
$b^2=\frac{16}{p^2}$
$-2ab=-2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right)=-2$

So, $\frac{p^2}{16}-2+\frac{16}{p^2}=\left(\frac{p}{4}-\frac{4}{p}\right)^2$ .

2(v)

9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc

Show solution

Compare $9a^2+4b^2+c^2-12ab+6ac-4bc$ with $(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$ .

Take $x=3a$ , $y=-2b$ , $z=c$ .
Then

$x^2=9a^2$
$y^2=4b^2$
$z^2=c^2$
$2xy=2(3a)(-2b)=-12ab$
$2yz=2(-2b)(c)=-4bc$
$2zx=2(c)(3a)=6ac$

So the factorisation is $(3a-2b+c)^2$ .

3(i)

(p + 3q + 7r)^2

Show solution

Use $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$ with $a=p$ , $b=3q$ , $c=7r$ .

So,

$(p+3q+7r)^2=p^2+(3q)^2+(7r)^2+2(p)(3q)+2(3q)(7r)+2(7r)(p)$

$=p^2+9q^2+49r^2+6pq+42qr+14pr$ .

3(ii)

(3x - 2y + 4z)^2

Show solution

Use $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$ with $a=3x$ , $b=-2y$ , $c=4z$ .

Then

$(3x-2y+4z)^2=(3x)^2+(-2y)^2+(4z)^2+2(3x)(-2y)+2(-2y)(4z)+2(4z)(3x)$

$=9x^2+4y^2+16z^2-12xy-16yz+24xz$ .

4Is this an identity?

(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2.

Show solution

Expand each square and add:

$(a+b-c)^2=a^2+b^2+c^2+2ab-2ac-2bc$

$(a-b+c)^2=a^2+b^2+c^2-2ab+2ac-2bc$

$(a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc$

Adding them gives:

$3a^2+3b^2+3c^2$

So the left side equals $2a^2+2b^2+2c^2$ only if the statement is adjusted? Wait, the sum actually is $3a^2+3b^2+3c^2$ , not $2a^2+2b^2+2c^2$ .

Hence the given statement is not an identity.

1(i)

35^2

Show solution

Use $(a-b)^2=a^2-2ab+b^2$ :

$35^2=(30+5)^2$ or $(40-5)^2$ .
Using $(30+5)^2$ :

$35^2=30^2+2(30)(5)+5^2=900+300+25=1225$ .

1(ii)

65^2

Show solution

Use $(a-b)^2=a^2-2ab+b^2$ :

$65^2=(60+5)^2=60^2+2(60)(5)+5^2=3600+600+25=4225$ .

1(iii)

85^2

Show solution

Use $(a+b)^2=a^2+2ab+b^2$ :

$85^2=(80+5)^2=80^2+2(80)(5)+5^2=6400+800+25=7225$ .

1(iv)

105^2

Show solution

Use $(a+b)^2=a^2+2ab+b^2$ :

$105^2=(100+5)^2=100^2+2(100)(5)+5^2=10000+1000+25=11025$ .

2Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.Show solution

The two rows of figures represent the expansion of a square of side $a+b+c$ into smaller squares and rectangles. So the identity is

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca.$

4Is this an identity?Show solution

Expand the left side:

$(a+b-c)^2+(a-b+c)^2+(a-b-c)^2$

$=(a^2+b^2+c^2+2ab-2ac-2bc)$
$+(a^2+b^2+c^2-2ab+2ac-2bc)$
$+(a^2+b^2+c^2-2ab-2ac+2bc)$

$=3a^2+3b^2+3c^2-2ab-2ac-2bc$

So it does not equal $2a^2+2b^2+2c^2$ for all values. Therefore, it is not an identity.

1(i)Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.Show solution

Compute each square using the most suitable identity:

$117^2=(100+17)^2=10000+3400+289=13689$
$78^2=(80-2)^2=6400-320+4=6084$
$198^2=(200-2)^2=40000-800+4=39204$
$214^2=(200+14)^2=40000+5600+196=45796$
$1104^2=(1100+4)^2=1210000+8800+16=1218816$
$1120^2=(1100+20)^2=1210000+44000+400=1254400$

For these, the square of a sum or square of a difference identities are easiest, depending on which nearby round number is chosen.

Exercise Set 4.4

1(i)s^2 - 11s + 24 = (\underline{\hspace{2cm}}) (\underline{\hspace{2cm}})Show solution

Find two numbers whose product is $24$ and sum is $-11$ . They are $-3$ and $-8$ .

So,

$s^2 - 11s + 24 = s^2 - 3s - 8s + 24 = s(s-3) - 8(s-3) = (s-3)(s-8).$

1(ii)(\underline{\hspace{2cm}})(x + 1) = (3x^2 - 4x - 7)Show solution

Let the missing factor be $A$ . Then

$A(x+1)=3x^2-4x-7.$

Factor by grouping:

$3x^2-4x-7=3x^2+3x-7x-7=3x(x+1)-7(x+1)=(3x-7)(x+1).$

So the missing factor is $3x-7$ .

1(iii)10x^2 - 11x - 6 = (2x - \underline{\hspace{2cm}}) (\underline{\hspace{2cm}} + 2)Show solution

We need

$(2x-\_)(\_+2)=10x^2-11x-6.$

Try splitting the middle term by comparing with $(2x-a)(5x+2)$ :

$(2x-a)(5x+2)=10x^2+4x-5ax-2a = 10x^2+(4-5a)x-2a.$

Match coefficients with $10x^2-11x-6$ :

$-2a=-6 \Rightarrow a=3,$

and then

$4-5(3)=4-15=-11.$

So,

$10x^2-11x-6=(2x-3)(5x+2).$

1(iv)6x^2 + 7x + 2 = (\underline{\hspace{2cm}}) (\underline{\hspace{2cm}})Show solution

Factor $6x^2+7x+2$ by splitting the middle term.

We need two numbers whose product is $6\cdot 2=12$ and sum is $7$ . They are $3$ and $4$ .

$6x^2+7x+2=6x^2+3x+4x+2=3x(2x+1)+2(2x+1)=(3x+2)(2x+1).$

2(i)

(41)^2

Show solution

Use $(a+b)^2$ :

$41^2=(40+1)^2=40^2+2\cdot40\cdot1+1^2=1600+80+1=1681.$

2(ii)

(27)^2

Show solution

Use $(a-b)^2$ :

$27^2=(30-3)^2=30^2-2\cdot30\cdot3+3^2=900-180+9=729.$

2(iii)

(23 \times 17)

Show solution

Use distributive property:

$23\times17=23\times(20-3)=460-69=391.$

2(iv)

(135)^2

Show solution

Use $(a-b)^2$ :

$135^2=(140-5)^2=140^2-2\cdot140\cdot5+5^2=19600-1400+25=18225.$

2(v)

(97)²

Show solution

Use $(a-b)^2$ :

$97^2=(100-3)^2=100^2-2\cdot100\cdot3+3^2=10000-600+9=9409.$

2(vi)

(18 × 29)

Show solution

Use distributive property:

$18\times29=18\times(30-1)=540-18=522.$

2(vii)

(34 × 43)

Show solution

Use distributive property:

$34\times43=34\times(40+3)=1360+102=1462.$

2(viii)

(205)²

Show solution

Use $(a+b)^2$ :

$205^2=(200+5)^2=200^2+2\cdot200\cdot5+5^2=40000+2000+25=42025.$

3(i)9a² + b² + 4c² - 6ab + 12ac - 4bcShow solution

Use the identity $(a-b)^2=a^2-2ab+b^2$ .

Here,

$9a^2=(3a)^2,\quad b^2=b^2,\quad 4c^2=(2c)^2.$

Check the middle terms:

$-6ab+12ac-4bc = -2(3a)(b)+2(3a)(2c)-2(b)(2c).$

So the expression is a perfect square:

$9a^2+b^2+4c^2-6ab+12ac-4bc=(3a-b+2c)^2.$

3(ii)16s² + 25t² - 40st

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3(iii)r² - r - 42

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3(iv)49g² + 14gh + h²

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3(v)64u² + 121v² + 4w² - 176uv - 32uw + 44vw

Group the expression as a perfect square of three terms:

$64u^2+121v^2+4w^2-176uv-32uw+44vw$

$=(8u)^2+(11v)^2+(2w)^2-2(8u)(11v)-2(8u)(2w)+2(11v)(2w).$

This matches

$(a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc$

with $a=8u$ , $b=11v$ , $c=2w$ .

So,

$64u^2+121v^2+4w^2-176uv-32uw+44vw=(8u-11v-2w)^2.$

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1(i)Fill in the blanks to complete the following identities:

s^2 - 11s + 24 = (\underline{\hspace{2cm}}) (\underline{\hspace{2cm}})

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1(ii)Fill in the blanks to complete the following identities:

(\underline{\hspace{2cm}})(x + 1) = (3x^2 - 4x - 7)

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1(iii)Fill in the blanks to complete the following identities:

10x^2 - 11x - 6 = (2x - \underline{\hspace{2cm}}) (\underline{\hspace{2cm}} + 2)

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1(iv)Fill in the blanks to complete the following identities:

6x^2 + 7x + 2 = (\underline{\hspace{2cm}}) (\underline{\hspace{2cm}})

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1(i)Fill in the blanks to complete the following identities:

Using the identities from the chapter, we can complete the standard forms:

$(x+y)^2 = x^2+2xy+y^2$
$(x-y)^2 = x^2-2xy+y^2$
$(x+y)(x-y)=x^2-y^2$
$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$
$(ax+b)(cx+d)=acx^2+(ad+bc)x+bd$
$x^3-y^3=(x-y)(x^2+xy+y^2)$
$x^3+y^3=(x+y)(x^2-xy+y^2)$

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Exercise Set 4.5

1(i)

\frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}

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1(ii)

\frac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}

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1(iii)\frac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}

Use the identity from the chapter:

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz).$

Here the denominator is

$w^2+v^2+x^2-2wv-2vx+2wx,$

which matches the identity after suitable arrangement in the chapter’s example, and the numerator is the corresponding cubic form. So the expression simplifies to the sum of the three variables:

$\frac{w^3-v^3+x^3+3wvx}{w^2+v^2+x^2-2wv-2vx+2wx}=w+v+x.$

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1(iv)

\frac{4y^2 - 20yz + 25z^2}{(25z^2 - 4y^2)}

Factor the numerator and denominator:

$4y^2-20yz+25z^2=(2y-5z)^2.$

Also,

$25z^2-4y^2=(5z-2y)(5z+2y).$

So,

$\frac{4y^2-20yz+25z^2}{25z^2-4y^2}=\frac{(2y-5z)^2}{(5z-2y)(5z+2y)}.$

Since $2y-5z=-(5z-2y)$ , one factor cancels with a minus sign:

$=\frac{-(2y-5z)}{5z+2y}=\frac{5z-2y}{5z+2y}.$

Thus the simplified form is $\frac{5z-2y}{5z+2y}.$

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1(v)

\frac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}

Factor each polynomial:

$x^2+x-6=(x+3)(x-2)$

$x^2-7x+12=(x-3)(x-4)$

$x^2-6x+8=(x-2)(x-4)$

$x^2-9=(x-3)(x+3)$

Now substitute and cancel common factors:

$\frac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)} =\frac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)}=1.$

So the expression simplifies to $1$ .

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1(vi)

\frac{p^4 - 16}{p^2 - 4p + 4}

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3Find the length and breadth of the pool.

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4(i)

(-3x + 4)^2

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4(ii)(2s + 7)(2s - 7)

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4(iii)

\left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right)

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4(iv)(2n + 7)(2n - 7)

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4(v)

(s - 2t)(s^2 + 2st + 4t^2)

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4(vi)

\left(\frac{1}{2r} - 4r\right)^2

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4(vii)

(-3m + 4k - l)^2

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4(viii)

\left(x - \frac{1}{3}y\right)^3

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4(ix)

\left(\frac{7}{2}k - \frac{2}{3}m\right)^3

Use $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ with $a=\frac{7}{2}k$ and $b=\frac{2}{3}m$ :

$\left(\frac{7}{2}k-\frac{2}{3}m\right)^3$

$=\left(\frac{7}{2}k\right)^3-3\left(\frac{7}{2}k\right)^2\left(\frac{2}{3}m\right)+3\left(\frac{7}{2}k\right)\left(\frac{2}{3}m\right)^2-\left(\frac{2}{3}m\right)^3$

$=\frac{343}{8}k^3-49k^2m+\frac{14}{3}km^2-\frac{8}{27}m^3.$

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5(i)

17 \times 21

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5(ii)

104 \times 96

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5(iii)

24 \times 16

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5(iv)

147^3

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5(v)

199^3

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5(vi)

127^3

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5(vii)

(-107)^3

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5(viii)

(-299)^3

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6(i)

4x^2 + 4x + 1

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6(ii)

9(3a^3 - 24b^3)

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6(iii)

s^3 + 125t^3

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7The village playground is shaped as a square of side 40 metres. A path of width

s

metres is created around the playground for people to walk. Find an expression for the area of the path in terms of

s

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8If a number plus its reciprocal equals

\frac{10}{3}

, find the number.

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9A rectangular pool has area

2x^2 + 7x + 3

square hastas. If its width is

2x + 1

hastas, find its length. Hasta was a unit used to measure length.

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10If both

x - 2

and

x - \frac{1}{2}

are factors of

px^2 + 5x + r

, show that

p = r

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11If

a + b + c = 5

and

ab + bc + ca = 10

, then prove that

a^3 + b^3 + c^3 - 3abc = -25

Use the identity

$(a+b+c)^3?$

Actually the chapter gives:

$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc$

First find $a^2+b^2+c^2$ from

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$

Given $a+b+c=5$ and $ab+bc+ca=10$ ,

$25=a^2+b^2+c^2+20$

$a^2+b^2+c^2=5$ .

Now

$a^2+b^2+c^2-ab-bc-ca=5-10=-5$ .

Therefore,

$a^3+b^3+c^3-3abc=(5)(-5)=-25$ .

So the result is $-25$ .

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12By factoring the expression, check that

n^3 - n

is always divisible by 6 for all natural numbers

n

. Give reasons.

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13(i)x^3 + y^3 - 12xy + 64, when

x + y = -4

Use the identity

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ .

Take $z=4$ because the expression is

$x^3+y^3+4^3-3xy(4)$

$=x^3+y^3+64-12xy$ .

Then $x+y=-4$ , so

$x^3+y^3+64-12xy=(x+y)(x^2+y^2+16-xy-4x-4y)$

A simpler direct use is:

For $x+y=-4$ ,

$(x+y)^3=x^3+y^3+3xy(x+y)$

$-64=x^3+y^3+3xy(-4)$

$-64=x^3+y^3-12xy$

Hence

$x^3+y^3-12xy+64=0$ .

So the value is 0.

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13(ii)x^3 - 8y^3 - 36xy - 216, when

x = 2y + 6

Given $x=2y+6$ , so

$x-2y=6$ .

The expression is

$x^3-8y^3-36xy-216=x^3-(2y)^3-3(x)(2y)(6)-6^3

But better, write it as

$x^3-(2y)^3-3x(2y)(6)-6^3$

and compare with

$a^3-b^3-3ab(a-b)?$

A cleaner approach: since $x=2y+6$ , set $x=2y+6$ in the expression:

$(2y+6)^3-8y^3-36(2y+6)y-216$

$=(2y+6)^3-(2y)^3-3(2y)(2y+6)(6)-6^3$

Using $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ , this simplifies to 0.

So the value is 0.

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1(i)Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:

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4(i)

\frac{4x^2 + 4x + 1}{4x^2 - 1}

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4(ii)

\frac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}

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4(iii)

\frac{s^3 + 125t^3}{s^2 - 2st - 35t^2}

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5(i)

25a^2 - 30ab + 9b^2

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5(ii)

36s^2 - 49t^2

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6(i)

6a^2 - 24b^2

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6(ii)

3ps^2 - 15ps + 12p

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13(i)

x^3 + y^3 - 12xy + 64

Use the identity from the chapter:

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ .

Here the expression is of the form

$x^3+y^3+4^3-3xy(4)$

which becomes 0 when $x+y=-4$ because

$x^3+y^3+64-12xy=(x+y)^3$ ? More directly:

Let $z=4$ . Then

$x^3+y^3+z^3-3xyz=x^3+y^3+64-12xy$ .

Since $x+y=-4$ , this expression is

$(x+y)^3+64= (-4)^3+64= -64+64=0$ .

So the value is 0.

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13(ii)

x^3 - 8y^3 - 36xy - 216

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Think and Reflect

1What can you say about

a

and

b

(a + b)^2 < a^2 + b^2

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2What can you say about

a

and

b

(a + b)^2 > a^2 + b^2

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3When will

(a + b)^2

be equal to

a^2 + b^2

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1Figure out the product of

x + 2

and

x + 3

using algebra tiles.

Using algebra tiles, $x+2$ is represented by one $x$ -tile and two unit tiles, and $x+3$ by one $x$ -tile and three unit tiles. Arranging them to form a rectangle gives:

$x \times x = x^2$
$x \times 3 = 3x$
$2 \times x = 2x$
$2 \times 3 = 6$

So the product is

$(x+2)(x+3)=x^2+3x+2x+6=x^2+5x+6.$

Thus the rectangle represents $x^2+5x+6$ , with sides $x+2$ and $x+3$ .

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6Suppose

7x

is split as

2x + 5x

; can a similar rectangular arrangement be formed? Consider other possibilities and check.

Yes. A similar rectangular arrangement can be formed if the split of $7x$ matches factors of the constant term. For example, in $x^2+7x+12$ , splitting $7x$ as $3x+4x$ works because $3\times 4=12$ . This gives a rectangle with sides $x+3$ and $x+4$ .

If we split $7x$ as $2x+5x$ , then we would need constants whose product is $12$ and sum is $7$ , but $2$ and $5$ give product $10$ , not $12$ . So this split does not form the required rectangle.

Other possibilities like $1x+6x$ also do not work for $12$ , because $1\times 6=6$ , not $12$ . So the correct rectangular arrangement depends on choosing numbers whose sum gives the middle term and whose product gives the constant term.

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7James and Reshma were talking about algebraic identities they learnt in school.

James: (a - b)² (a + b) = (a² - 2ab + b²)(a + b)

Reshma: I have a different idea. (a - b)² (a + b) = (a - b) [(a - b)(a + b)] = (a - b)(a² - b²)

I will find this product to get the answer.

According to you, who is correct and why?

Reshma is correct.

Starting with

$(a-b)^2(a+b),$

we can first use the identity

$(a-b)(a+b)=a^2-b^2.$

So,

$(a-b)^2(a+b)=(a-b)[(a-b)(a+b)]=(a-b)(a^2-b^2).$

Now expanding further,

$(a-b)(a^2-b^2)=a(a^2-b^2)-b(a^2-b^2)=a^3-ab^2-a^2b+b^3,$

which can also be written as

$a^3-a^2b-ab^2+b^3.$

James has only written the product in one factored form, while Reshma has used another identity to simplify it more usefully. So both start correctly, but Reshma’s method is the better simplification and shows the result more clearly.

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8Try to combine more such identities and find new results.

Using the identities from the chapter, we can combine them to get new results. For example:

From $x^2-y^2=(x-y)(x+y)$ and $x^3-y^3=(x-y)(x^2+xy+y^2)$ , we see that $x-y$ is a common factor of many expressions of the form $x^n-y^n$ .
Also,

$(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=x^3+y^3+z^3-3xyz.$

This gives a new identity.

From the cube identity,

$(a+b)^3=a^3+3a^2b+3ab^2+b^3,$

and replacing $b$ by $-b$ , we get

$(a-b)^3=a^3-3a^2b+3ab^2-b^3.$

So by combining known identities, we can generate new factorisations and expansions such as $x^3+y^3=(x+y)(x^2-xy+y^2)$ and $x^3-y^3=(x-y)(x^2+xy+y^2)$ .

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9What do you think (a + b)³ will look like?

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10Label the squares and rectangles in Fig. 4.4 so that it represents the identity

(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

To represent the identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca,$

label the figure like this:

the three squares as $a^2$ , $b^2$ , and $c^2$
the six rectangles as pairs contributing to the cross terms:
two rectangles of area $ab$ each, together giving $2ab$
two rectangles of area $bc$ each, together giving $2bc$
two rectangles of area $ca$ each, together giving $2ca$

So the whole square of side $a+b+c$ is partitioned into 3 squares and 6 rectangles, and their total area gives the identity.

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Frequently Asked Questions

What are the important topics in Exploring Algebraic Identities for CBSE Class 9 Mathematics?

Key topics in Exploring Algebraic Identities include Identity and Equation, Square of a Sum and Square of a Difference, Square of a Trinomial, Product of Binomials and Factorisation. Study these first, then practise questions on each for Class 9 exams.

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Exercise Set 4.1

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Exercise Set 4.5

Think and Reflect

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