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) $

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.