Simplify: $\frac{a^4+3a^2b^2+b^4-2a^3b-2ab^3}{a^4+a^2b^2+b^4}$

Question

Simplify, $$\cfrac{a^4+3a^2b^2+b^4-2a^3b-2ab^3}{a^4+a^2b^2+b^4}$$ Maybe this is basic, but I'm stuck!

Answer 1

The numerator $=(a^2+b^2)^2-2ab(a^2+b^2)+(ab)^2=(a^2+b^2-ab)^2$

The denominator $=(a^2+b^2)^2-(ab)^2=?$

Answer 2

We have $$\frac{a^4+3a^2b^2+b^4-2a^3b-2ab^3}{a^4+a^2b^2+b^4}$$ $$=\frac{(a^4+2a^2b^2+b^4)+a^2b^2-2a^3b-2ab^3}{(a^4+2a^2b^2+b^4)-a^2b^2}$$ $$=\frac{(a^2+b^2)^2-2ab(a^2+b^2)+a^2b^2}{(a^2+b^2)^2-a^2b^2}$$ $$=\frac{\overbrace{(a^2+b^2)^2-a^2b^2}\overbrace{-2ab(a^2+b^2)+2a^2b^2}}{(a^2+b^2-ab)(a^2+b^2+ab)}$$ $$=\frac{(a^2+b^2-ab)(a^2+b^2+ab)-2ab(a^2+b^2-ab)}{(a^2+b^2-ab)(a^2+b^2+ab)}$$ $$=\frac{(a^2+b^2-ab)[(a^2+b^2+ab)-2ab]}{(a^2+b^2-ab)(a^2+b^2+ab)}$$ $$=\frac{a^2+b^2-ab}{a^2+b^2+ab}$$