# Verify : (i) $x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)$(ii) $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)$

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To do:

We have to verify

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

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

Solution:

(i) We know that,

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

This implies,

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

Taking $(x+y)$ common, we get,

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

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

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

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

Hence verified.

(ii) We know that,

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

This implies,

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

Taking $(x-y)$ common, we get,

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

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

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

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

Hence verified.

Updated on 10-Oct-2022 13:39:07