## Precalculus (6th Edition)

$\color{blue}{\dfrac{x^2-xy+y^2}{x^2+xy+y^2}}$
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: $\require{cancel} \\=\dfrac{(x+y)(x^2-xy+y^2)}{(x-y)(x^2+xy+y^2)} \cdot \dfrac{(x-y)(x+y)}{(x+y)(x+y)} \\=\dfrac{\cancel{(x+y)}(x^2-xy+y^2)}{\cancel{(x-y)}(x^2+xy+y^2)} \cdot \dfrac{\cancel{(x-y)}\cancel{(x+y)}}{\cancel{(x+y)}\cancel{(x+y)}} \\=\color{blue}{\dfrac{x^2-xy+y^2}{x^2+xy+y^2}}$