NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4

1. Use suitable identities to find the following products:

1. (x+4)(x+10)

Ans: Using the identity, $$(x+a)(x+b)=x^2+(a+b)x+ab$$

Here we have, $$a=4,b=10$$

We get,
$$(x+4)(x+10)=x^2+(4+10)x+\left(4\right)\left(10\right)$$

$$=x^2+14x+40$$

2. (x+8)(x-10)

Ans: Using the identity, $$(x+a)(x+b)=x^2+(a+b)x+ab$$

Here we have, $$a=8,b=-10$$

We get,
$(x+8)(x+(-10))=x^2+(8+(-10))x+\left(8\right)(-10)$

 $(x+8)(x-10)=x^2+(8-10)x-80$

 $=x^2-2x-80$

3. ( 3x+4)( 3x-5)

Ans: Using the identity, $$(x+a)(x+b)=x^2+(a+b)x+ab$$

Here we have, $$a=4,b=-5$$

We get,
$(3x+4)(3x+(-5))={\left(3x\right)}^2+(4+(-5))3x+\left(4\right)(-5)$

$(3x+4)(3x-5)=9x^2+(4-5)3x-20$

$=9x^2-3x-20$

4. $\mathbf{(}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{)}\mathbf{(}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\frac{\mathbf{3}}{\mathbf{2}}\mathbf{)}$

Ans: Using the identity, $$(x+y)(x-y)=x^2-y^2$$

Here we have, $$x=y^2,y={\displaystyle \frac{3}{2}}$$

We get,
$(y^2+{\displaystyle \frac{3}{2}})(y^2-{\displaystyle \frac{3}{2}})={\left(y^2\right)}^2-{\left({\displaystyle \frac{3}{2}}\right)}^2$

$=y^4-{\displaystyle \frac{9}{4}}$

5. (3-2x)(3+2x)

Ans: Using the identity, $$(x+y)(x-y)=x^2-y^2$$

Here we have, $$x=3,y=2x$$

We get,
$(3+2x)(3-2x)={\left(3\right)}^2-{\left(2x\right)}^2$

$=9-4x^2$ 

2. Evaluate the following products without multiplying directly:

1. $$\mathbf{103}\times \mathbf{107}$$

Ans: $$103\times 107=(100+3)\times (100+7)$$

By using the identity, $$(x+a)(x+b)=x^2+(a+b)x+ab$$

Here we have, $$x=100,a=3,b=7$$

We get,

$(100+3)(100+7)={\left(100\right)}^2+(3+7)100+\left(3\right)\left(7\right)$

$\left(103\right)\times \left(107\right)=10000+1000+21$

$=11021$

2. $$\mathbf{95}\times \mathbf{96}$$

Ans: $$95\times 96=(100-5)\times (100-4)$$

By using the identity, $$(x-a)(x-b)=x^2-(a+b)x+ab$$

Here we have, $$x=100,a=5,b=4$$

We get,
$(100-5)(100-4)={\left(100\right)}^2-(5+4)100+\left(5\right)\left(4\right)$

$\left(95\right)\times \left(96\right)=10000-900+20$

$=9120$

3. $$\mathbf{104}\times \mathbf{96}$$

Ans: $$104\times 96=(100+4)\times (100-4)$$

By using the identity, $$(x+y)(x-y)=x^2-y^2$$

Here we have, $$x=100,y=4$$

We get,
$(100+4)(100-4)={\left(100\right)}^2-{\left(4\right)}^2$

$\left(104\right)\times \left(96\right)=10000-16$

$=9984$

3. Factorize the following using appropriate identities:

1. $$\mathbf{9}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{xy}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}$$

Ans: $$9x^2+6xy+y^2={\left(3x\right)}^2+2\left(3x\right)\left(y\right)+{\left(y\right)}^2$$

By using the identity, $$x^2+2xy+y^2={(x+y)}^2$$

Here, $$x=3x,y=y$$

$9x^2+6xy+y^2={\left(3x\right)}^2+2\left(3x\right)\left(y\right)+{\left(y\right)}^2$

$={(3x+y)}^2$

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

2. $$\mathbf{4}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{+}\mathbf{1}$$

Ans: $$4y^2-4y+1={\left(2y\right)}^2-2\left(2y\right)\left(1\right)+{\left(1\right)}^2$$

By using the identity, $$x^2-2xy+y^2={(x-y)}^2$$

Here, $$x=2y,y=1$$

$4y^2-4y+1={\left(2y\right)}^2-2\left(2y\right)\left(1\right)+{\left(1\right)}^2$ 

$={(2y-1)}^2$

$=(2y-1)(2y-1)$

3. $${\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\displaystyle \frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{100}}}$$

Ans: $$x^2-{\displaystyle \frac{y^2}{100}}={\left(x\right)}^2-{\left({\displaystyle \frac{y}{10}}\right)}^2$$

By using the identity, $$x^2-y^2=(x+y)(x-y)$$

Here, $$x=x,y={\displaystyle \frac{y}{10}}$$

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

$=(x-{\displaystyle \frac{y}{10}})(x+{\displaystyle \frac{y}{10}})$

4. Expand each of the following, using suitable identities:

1. $${(\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}\mathbf{4}\mathbf{z})}^{\mathbf{2}}$$

Ans: By using the identity, $${(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx$$

Here, $$x=x,y=2y,z=4z$$

${(x+2y+4z)}^2={\left(x\right)}^2+{\left(2y\right)}^2+{\left(4z\right)}^2+2\left(x\right)\left(2y\right)+2\left(2y\right)\left(4z\right)+2\left(4z\right)\left(x\right)$

$=x^2+4y^2+16z^2+4xy+16yz+8xz$

2. $${(\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{+}\mathbf{z})}^{\mathbf{2}}$$

Ans: By using the identity, $${(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx$$

Here, $$x=2x,y=-y,z=z$$

${(2x-y+z)}^2={\left(2x\right)}^2+{(-y)}^2+{\left(z\right)}^2+2\left(2x\right)(-y)+2(-y)\left(z\right)+2\left(z\right)\left(2x\right)$

 $=4x^2+y^2+z^2-4xy-2yz+4xz$

3. $${(\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{z})}^{\mathbf{2}}$$

Ans: By using the identity, $${(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx$$

Here, $$x=-2x,y=3y,z=2z$$

${(-2x+3y+2z)}^2={(-2x)}^2+{\left(3y\right)}^2+{\left(2z\right)}^2+2(-2x)\left(3y\right)+2\left(3y\right)\left(2z\right)+2\left(2z\right)(-2x)$

$=4x^2+9y^2+4z^2-12xy+12yz-8xz$

4. $${(\mathbf{3}\mathbf{a}\mathbf{-}\mathbf{7}\mathbf{b}\mathbf{-}\mathbf{c})}^{\mathbf{2}}$$

Ans: By using the identity, $${(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx$$

Here, $$x=3a,y=-7b,z=-c$$

${(3a-7b-c)}^2={\left(3a\right)}^2+{(-7b)}^2+{(-c)}^2+2\left(3a\right)(-7b)+2(-7b)(-c)+2(-c)\left(3a\right)$

$=9a^2+49b^2+c^2-42ab+14bc-6ca$

5. $${(\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{5}\mathbf{y}\mathbf{-}\mathbf{3}\mathbf{z})}^{\mathbf{2}}$$

Ans: By using the identity, $${(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx$$

Here, $$x=-2x,y=5y,z=-3z$$

${(-2x+5y-3z)}^2={(-2x)}^2+{\left(5y\right)}^2+{(-3z)}^2+2(-2x)\left(5y\right)+2\left(5y\right)(-3z)+2(-3z)(-2x)$

$=4x^2+25y^2+9z^2-20xy-30yz+12xz$

6. $${({\displaystyle \frac{\mathbf{1}}{\mathbf{4}}}\mathbf{a}\mathbf{-}{\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}\mathbf{b}\mathbf{+}\mathbf{1})}^{\mathbf{2}}$$

Ans: By using the identity, $${(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx$$

Here, $$x={\displaystyle \frac{1}{4}}a,y=-{\displaystyle \frac{1}{2}}b,z=1$$

${({\displaystyle \frac{1}{4}}a-{\displaystyle \frac{1}{2}}b+1)}^2={\left({\displaystyle \frac{1}{4}}a\right)}^2+{(-{\displaystyle \frac{1}{2}}b)}^2+{\left(1\right)}^2+2\left({\displaystyle \frac{1}{4}}a\right)(-{\displaystyle \frac{1}{2}}b)+2(-{\displaystyle \frac{1}{2}}b)\left(1\right)+2\left(1\right)\left({\displaystyle \frac{1}{4}}a\right)$

$={\displaystyle \frac{1}{16}}{a}^2+{\displaystyle \frac{1}{4}}{b}^2+1-{\displaystyle \frac{1}{4}}ab-b+{\displaystyle \frac{1}{2}}a$

5. Factorise:

1. $$\mathbf{4}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{9}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{16}{\mathbf{z}}^{\mathbf{2}}\mathbf{+}\mathbf{12}\mathbf{xy}\mathbf{-}\mathbf{24}\mathbf{yz}\mathbf{-}\mathbf{16}\mathbf{xz}$$

Ans: By using the identity, $${(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx$$

We can see that, $$x^2+y^2+z^2+2xy+2yz+2zx={(x+y+z)}^2$$

$4x^2+9y^2+16z^2+12xy-24yz-16xz={\left(2x\right)}^2+{\left(3y\right)}^2+{(-4z)}^2+2\left(2x\right)\left(3y\right)+2\left(3y\right)(-4z)+2(-4z)\left(2x\right)$

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

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

2. $$\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}\mathbf{8}{\mathbf{z}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\sqrt{\mathbf{2}}\mathbf{xy}\mathbf{+}\mathbf{4}\sqrt{\mathbf{2}}\mathbf{yz}\mathbf{-}\mathbf{8}\mathbf{xz}$$

Ans: By using the identity, $${(x+y+z)}^2=x^2+y^2+z^2+2xy+2yz+2zx$$

We can see that, $$x^2+y^2+z^2+2xy+2yz+2zx={(x+y+z)}^2$$

$2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz={(-\sqrt{2}x)}^2+{\left(y\right)}^2+{\left(2\sqrt{2}z\right)}^2+2(-\sqrt{2}x)\left(y\right)+2\left(y\right)\left(2\sqrt{2}z\right)+2\left(2\sqrt{2}z\right)(-\sqrt{2}x)$

$={(-\sqrt{2}x+y-2\sqrt{2}z)}^2$

$=(-\sqrt{2}x+y-2\sqrt{2}z)(-\sqrt{2}x+y-2\sqrt{2}z)$

6. Write the following cubes in expanded form:

1. $${(\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{1})}^{\mathbf{3}}$$

Ans: By using the identity, $${(x+y)}^3=x^3+y^3+3xy(x+y)$$

${(2x+1)}^3={\left(2x\right)}^3+{\left(1\right)}^3+3\left(2x\right)\left(1\right)(2x+1)$

$=8x^3+1+6x(2x+1)$

$=8x^3+1+12x^2+6x$

$=8x^3+12x^2+6x+1$

2. $${(\mathbf{2}\mathbf{a}\mathbf{-}\mathbf{3}\mathbf{b})}^{\mathbf{3}}$$

Ans: By using the identity, $${(x-y)}^3=x^3-y^3-3xy(x-y)$$

${(2a-3b)}^3={\left(2a\right)}^3-{\left(3b\right)}^3-3\left(2a\right)\left(3b\right)(2a-3b)$

$=8x^3-27b^3-18ab(2a-3b)$

$=8x^3-27b^3-36a^{2}b+54a{b}^2$

3. $${({\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}\mathbf{x}\mathbf{+}\mathbf{1})}^{\mathbf{3}}$$

Ans: By using the identity, $${(x+y)}^3=x^3+y^3+3xy(x+y)$$

${({\displaystyle \frac{3}{2}}x+1)}^3={\left({\displaystyle \frac{3}{2}}x\right)}^3+{\left(1\right)}^3+3\left({\displaystyle \frac{3}{2}}x\right)\left(1\right)({\displaystyle \frac{3}{2}}x+1)$ 

$={\displaystyle \frac{27}{8}}{x}^3+1+{\displaystyle \frac{9}{2}}x({\displaystyle \frac{3}{2}}x+1)$

$={\displaystyle \frac{27}{8}}{x}^3+1+{\displaystyle \frac{27}{4}}{x}^2+{\displaystyle \frac{9}{2}}x$ 

$={\displaystyle \frac{27}{8}}{x}^3+{\displaystyle \frac{27}{4}}{x}^2+{\displaystyle \frac{9}{2}}x+1$

4. $${(\mathbf{x}\mathbf{-}{\displaystyle \frac{\mathbf{2}}{\mathbf{3}}}\mathbf{y})}^{\mathbf{3}}$$

Ans: By using the identity, $${(x-y)}^3=x^3-y^3-3xy(x-y)$$

${(x-{\displaystyle \frac{2}{3}}y)}^3={\left(x\right)}^3-{\left({\displaystyle \frac{2}{3}}y\right)}^3-3\left(x\right)\left({\displaystyle \frac{2}{3}}y\right)(x-{\displaystyle \frac{2}{3}}y)$

$=x^3-{\displaystyle \frac{8}{27}}{y}^3-2xy(x-{\displaystyle \frac{2}{3}}y)$

$=x^3-{\displaystyle \frac{8}{27}}{y}^3-2x^{2}y+{\displaystyle \frac{4}{3}}x{y}^2$

7. Evaluate the following using suitable identities:

1. $${\left(\mathbf{99}\right)}^{\mathbf{3}}$$

Ans: Here we can write $${\left(99\right)}^3$$ as $${(100-1)}^3$$

By using the identity, $${(x-y)}^3=x^3-y^3-3xy(x-y)$$

${(100-1)}^3={\left(100\right)}^3-{\left(1\right)}^3-3\left(100\right)\left(1\right)(100-1)$

$=1000000-1-300(100-1)$

$=1000000-1-30000+300$

$=970299$

2. $${\left(\mathbf{102}\right)}^{\mathbf{3}}$$

Ans: Here we can write $${\left(102\right)}^3$$ as $${(100+2)}^3$$

By using the identity, $${(x+y)}^3=x^3+y^3+3xy(x+y)$$

${(100+2)}^3={\left(100\right)}^3+{\left(2\right)}^3+3\left(100\right)\left(2\right)(100+2)$

$=1000000+8+600(100+2)$

$=1000000+8+60000+1200$

$=1061208$

3. $${\left(\mathbf{998}\right)}^{\mathbf{3}}$$

Ans: Here we can write $${\left(998\right)}^3$$ as $${(1000-2)}^3$$

By using the identity, $${(x-y)}^3=x^3-y^3-3xy(x-y)$$

${(1000-2)}^3={\left(1000\right)}^3-{\left(2\right)}^3-3\left(1000\right)\left(2\right)(1000-2)$

 $=1000000000-8-6000(1000-2)$

 $=1000000000-8-6000000+12000$

 $=994011992$

8. Factorise each of the following:

1. $$\mathbf{8}{\mathbf{a}}^{\mathbf{3}}\mathbf{+}{\mathbf{b}}^{\mathbf{3}}\mathbf{+}\mathbf{12}{\mathbf{a}}^{\mathbf{2}}\mathbf{b}\mathbf{+}\mathbf{6}\mathbf{a}{\mathbf{b}}^{\mathbf{2}}$$

Ans: Here we can write $$8a^3+b^3+12a^{2}b+6a{b}^2$$ as 

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

By using the identity, $${(x+y)}^3=x^3+y^3+3xy(x+y)$$

Here, $$x=2a,y=b$$

$8a^3+b^3+12a^{2}b+6a{b}^2={\left(2a\right)}^3+{\left(b\right)}^3+3{\left(2a\right)}^2\left(b\right)+3\left(2a\right){\left(b\right)}^2$

 $={(2a+b)}^3$

 $=(2a+b)(2a+b)(2a+b)$

2. $$\mathbf{8}{\mathbf{a}}^{\mathbf{3}}\mathbf{-}{\mathbf{b}}^{\mathbf{3}}\mathbf{-}\mathbf{12}{\mathbf{a}}^{\mathbf{2}}\mathbf{b}\mathbf{+}\mathbf{6}\mathbf{a}{\mathbf{b}}^{\mathbf{2}}$$

Ans: Here we can write $$8a^3-b^3-12a^{2}b+6a{b}^2$$ as 

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

By using the identity, $${(x-y)}^3=x^3-y^3-3xy(x-y)$$

Here, $$x=2a,y=b$$

$8a^3-b^3-12a^{2}b+6a{b}^2={\left(2a\right)}^3-{\left(b\right)}^3-3{\left(2a\right)}^2\left(b\right)+3\left(2a\right){\left(b\right)}^2$

 $={(2a-b)}^3$

 $=(2a-b)(2a-b)(2a-b)$

3. $$\mathbf{27}\mathbf{-}\mathbf{125}{\mathbf{a}}^{\mathbf{3}}\mathbf{-}\mathbf{135}\mathbf{a}\mathbf{+}\mathbf{225}{\mathbf{a}}^{\mathbf{2}}$$

Ans: Here we can write $$27-125a^3-135a+225a^2$$ as 

$${\left(3\right)}^3-{\left(5a\right)}^3-3{\left(3\right)}^2\left(5a\right)+3\left(3\right){\left(5a\right)}^2$$

By using the identity, $${(x-y)}^3=x^3-y^3-3xy(x-y)$$

Here, $$x=3,y=5a$$

$27-125a^3-135a+225a^2={\left(3\right)}^3-{\left(5a\right)}^3-3{\left(3\right)}^2\left(5a\right)+3\left(3\right){\left(5a\right)}^2$

 $={(3-5a)}^3$

 $=(3-5a)(3-5a)(3-5a)$

4. $$\mathbf{64}{\mathbf{a}}^{\mathbf{3}}\mathbf{-}\mathbf{27}{\mathbf{b}}^{\mathbf{3}}\mathbf{-}\mathbf{144}{\mathbf{a}}^{\mathbf{2}}\mathbf{b}\mathbf{+}\mathbf{108}\mathbf{a}{\mathbf{b}}^{\mathbf{2}}$$

Ans: Here we can write $$64a^3-27b^3-144a^{2}b+108a{b}^2$$ as 

$${\left(4a\right)}^3-{\left(3b\right)}^3-3{\left(4a\right)}^2\left(3b\right)+3\left(4a\right){\left(3b\right)}^2$$

By using the identity, $${(x-y)}^3=x^3-y^3-3xy(x-y)$$

Here, $$x=4a,y=3b$$

$64a^3-27b^3-144a^{2}b+108a{b}^2={\left(4a\right)}^3-{\left(3b\right)}^3-3{\left(4a\right)}^2\left(3b\right)+3\left(4a\right){\left(3b\right)}^2$ 

$={(4a-3b)}^3$

$=(4a-3b)(4a-3b)(4a-3b)$

5. $$\mathbf{27}{\mathbf{p}}^{\mathbf{3}}\mathbf{-}{\displaystyle \frac{\mathbf{1}}{\mathbf{216}}}\mathbf{-}{\displaystyle \frac{\mathbf{9}}{\mathbf{2}}}{\mathbf{p}}^{\mathbf{2}}\mathbf{+}{\displaystyle \frac{\mathbf{1}}{\mathbf{4}}}\mathbf{p}$$

Ans: Here we can write $$27p^3-{\displaystyle \frac{1}{216}}-{\displaystyle \frac{9}{2}}{p}^2+{\displaystyle \frac{1}{4}}p$$ as 

$${\left(3p\right)}^3-{\left({\displaystyle \frac{1}{6}}\right)}^3-3{\left(3p\right)}^2\left({\displaystyle \frac{1}{6}}\right)+3\left(3p\right){\left({\displaystyle \frac{1}{6}}\right)}^2$$

By using the identity, $${(x-y)}^3=x^3-y^3-3xy(x-y)$$

Here, $$x=3p,y={\displaystyle \frac{1}{6}}$$

$27p^3-{\displaystyle \frac{1}{216}}-{\displaystyle \frac{9}{2}}{p}^2+{\displaystyle \frac{1}{4}}p={\left(3p\right)}^3-{\left({\displaystyle \frac{1}{6}}\right)}^3-3{\left(3p\right)}^2\left({\displaystyle \frac{1}{6}}\right)+3\left(3p\right){\left({\displaystyle \frac{1}{6}}\right)}^2$

$={(3p-{\displaystyle \frac{1}{6}})}^3$

$=(3p-{\displaystyle \frac{1}{6}})(3p-{\displaystyle \frac{1}{6}})(3p-{\displaystyle \frac{1}{6}})$

9. Verify:

1. $${\mathbf{x}}^{\mathbf{3}}\mathbf{+}{\mathbf{y}}^{\mathbf{3}}\mathbf{=}(\mathbf{x}\mathbf{+}\mathbf{y})({\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{xy}\mathbf{+}{\mathbf{y}}^{\mathbf{2}})$$

Ans: By using the identity, $${(x+y)}^3=x^3+y^3+3xy(x+y)$$

$$x^3+y^3={(x+y)}^3-3xy(x+y)$$

$$x^3+y^3=(x+y)[{(x+y)}^2-3xy]$$Taking $$(x+y)$$ common

$x^3+y^3=(x+y)[(x^2+y^2+2xy)-3xy]$

$\Rightarrow x^3+y^3=(x+y)(x^2+y^2-xy)$

Hence, verified.

2. $${\mathbf{x}}^{\mathbf{3}}\mathbf{-}{\mathbf{y}}^{\mathbf{3}}\mathbf{=}(\mathbf{x}\mathbf{-}\mathbf{y})({\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{xy}\mathbf{+}{\mathbf{y}}^{\mathbf{2}})$$

Ans: By using the identity, $${(x-y)}^3=x^3-y^3-3xy(x-y)$$

$$x^3-y^3={(x-y)}^3+3xy(x+y)$$

$$x^3-y^3=(x-y)[{(x-y)}^2+3xy]$$

Taking $$(x-y)$$ common

$x^3-y^3=(x-y)[(x^2+y^2-2xy)+3xy]$

$\Rightarrow x^3-y^3=(x-y)(x^2+y^2+xy)$

Hence, verified.