Question: Simplify: $\dfrac{x-y}{x^2-xy+y^2} +$$ \dfrac{x+y}{x^2+xy+y^2} +$$ \dfrac{2y^3}{x^4+x^2y^2+y^4}$

Solution:
Given,

$= \dfrac{x-y}{x^2-xy+y^2} + \dfrac{x+y}{x^2+xy+y^2} + \dfrac{2y^3}{x^4+x^2y^2+y^4}$

$= \dfrac{x-y}{(x^2+y^2) - (xy)} + \dfrac{x+y}{(x^2+y^2)+(xy)} + \dfrac{2y^3}{x^4+x^2y^2+y^4}$

$= \dfrac{(x-y)(x^2+y^2+xy) + (x+y)(x^2+y^2-xy)}{\{(x^2+y^2)-(xy)\}\{(x^2+y^2)+(xy)\}} + \dfrac{2y^3}{x^4+x^2y^2+y^4}$

$= \dfrac{x^3 + xy^2 +x^2y - x^2y -y^3-xy^2 + x^3 +xy^2 -x^2y +x^2y+y^3-xy^2}{(x^2+y^2)^2 - (xy)^2} + \dfrac{2y^3}{x^4+x^2y^2+y^4}$

$= \dfrac{2x^3}{x^4 +2x^2y^2 + y^4 -x^2y^2} + \dfrac{2y^3}{x^4+x^2y^2+y^4}$

[ We know, x^4+x^2y^2+y^4 = (x²+xy+y²)(x²-xy+y²) ]

$= \dfrac{2x^3}{x^4+x^2y^2+y^4} + \dfrac{2y^3}{x^4+x^2y^2+y^4}$

$=\dfrac{2x³+2y³}{x^4+x^2y^2+y^4}$

$= \dfrac{2(x³ +y³)}{x^4+x^2y^2+y^4}$

$= \dfrac{2(x+y)(x²-xy+y²)}{(x²+xy+y²)(x²-xy+y²)}$

$= \dfrac{2(x+y)}{x²+xy+y²}$

= Answer