Question

Упростите:
(1/(a-b)-(a^2+ab)/(a^3-b^3))/((b^2)/(a^2+ab+b^2))

Answer 1

$(\frac{1}{a-b}-\frac{a^2+ab}{a^3-b^3}):\frac{b^2}{a^2+ab+b^2}=(\frac{1}{a-b}-\frac{a^2+ab}{(a-b)(a^2+ab+b^2)}):\frac{b^2}{a^2+ab+b^2}=(\frac{a^2+ab+b^2}{(a-b)(a^2+ab+b^2)}-\frac{a^2+ab}{(a-b)(a^2+ab+b^2)}):\frac{b^2}{a^2+ab+b^2}=(\frac{a^2+ab+b^2-a^2-ab}{(a-b)(a^2+ab+b^2)}):\frac{b^2}{a^2+ab+b^2}=\frac{b^2}{(a-b)(a^2+ab+b^2)}}\cdot\frac{a^2+ab+b^2}{b^2}=\frac{b^2(a^2+ab+b^2)}{(a-b)(a^2+ab+b^2)b^2}}=\frac{1}{a-b}}$

Answer 2

$(\frac{1}{a-b}-\frac{a^2+ab}{a^3-b^3}):\frac{b^2}{a^2+ab+b^2}=\frac{a^2+ab+b^2-a^2-ab}{(a-b)(a^2+ab+b^2)}\cdot \frac{a^2+ab+b^2}{b^2}=\\\\=\frac{b^2}{(a-b)(a^2+ab+b^2)}\cdot \frac{a^2+ab+b^2}{b^2}=\frac{1}{a-b}$