Answer
$\color{blue}{\dfrac{x^2-xy+y^2}{x^2+xy+y^2}}$
Work Step by Step
RECALL:
(1) $a^2-b^2=(a-b)(a+b)$
(2) $a^3-b^3=(a-b)(a^2+ab+b^2)$
(3) $a^3+b^3=(a+b)(a^2-ab+b^2)$
(4) $a^2+2ab+b^2=(a+b)(a+b)$
Factor each polynomial using the formula above, then cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{(x+y)(x^2-xy+y^2)}{(x-y)(x^2+xy+y^2)} \cdot \dfrac{(x-y)(x+y)}{(x+y)(x+y)}
\\=\dfrac{\cancel{(x+y)}(x^2-xy+y^2)}{\cancel{(x-y)}(x^2+xy+y^2)} \cdot \dfrac{\cancel{(x-y)}\cancel{(x+y)}}{\cancel{(x+y)}\cancel{(x+y)}}
\\=\color{blue}{\dfrac{x^2-xy+y^2}{x^2+xy+y^2}}$
