Question

39.21 (1) Даю 50 баллов

Answer 1

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