Question

Помогите упростить пожалуйста

Answer 1

$mathtt{frac{a+b}{a^{frac{5}{3}}b^{frac{5}{3}}}*frac{a^2b^{frac{5}{3}}-a^{frac{5}{3}}b^2}{a^{frac{2}{3}}-a^{frac{1}{3}}b^{frac{1}{3}}+b^{frac{2}{3}}}=frac{a+b}{a^{frac{5}{3}}b^{frac{5}{3}}}*frac{a^{frac{5}{3}}b^{frac{5}{3}}(a^{frac{1}{3}}-b^{frac{1}{3}})}{a^{frac{2}{3}}-a^{frac{1}{3}}b^{frac{1}{3}}+b^{frac{2}{3}}}=frac{(a+b)(a^{frac{1}{3}}-b^{frac{1}{3}})}{a^{frac{2}{3}}-a^{frac{1}{3}}b^{frac{1}{3}}+b^{frac{2}{3}}}=}$$mathtt{\\frac{(a+b)(a^{frac{1}{3}}-b^{frac{1}{3}})(a^{frac{1}{3}}+b^{frac{1}{3}})}{(a^{frac{2}{3}}-a^{frac{1}{3}}b^{frac{1}{3}}+b^{frac{2}{3}})(a^{frac{1}{3}}+b^{frac{1}{3}})}=frac{(a+b)[(a^{frac{1}{3}})^2-(b^{frac{1}{3}})^2]}{(a^{frac{1}{3}})^3+(b^{frac{1}{3}})^3}=frac{(a+b)(a^{frac{2}{3}}-b^{frac{2}{3}})}{a+b}=sqrt[3]{mathtt{a^2}}-sqrt[3]{mathtt{b^2}}}$


$mathtt{(frac{7}{b^{frac{1}{6}}+7}+frac{b^{frac{1}{3}}+34}{b^{frac{1}{3}}-49}-frac{7}{b^{frac{1}{6}}-7}):frac{b^{frac{1}{6}}-8}{(b^{frac{1}{6}}-7)^2}=frac{7(b^{frac{1}{6}}-7)+b^{frac{1}{3}}+34-7(b^{frac{1}{6}}+7)}{(b^{frac{1}{6}}+7)(b^{frac{1}{6}}-7)}:frac{b^{frac{1}{6}}-8}{(b^{frac{1}{6}}-7)^2}=}$$mathtt{\\frac{7b^{frac{1}{6}}-49+b^{frac{1}{3}}+34-7b^{frac{1}{6}}-49}{(b^{frac{1}{6}}+7)(b^{frac{1}{6}}-7)}*frac{(b^{frac{1}{6}}-7)^2}{b^{frac{1}{6}}-8}=frac{b^{frac{1}{3}}-64}{b^{frac{1}{6}}+7}*frac{b^{frac{1}{6}}-7}{b^{frac{1}{6}}-8}=frac{(b^{frac{1}{6}}-8)(b^{frac{1}{6}}+8)}{b^{frac{1}{6}}+7}*frac{b^{frac{1}{6}}-7}{b^{frac{1}{6}}-8}=}\\mathtt{frac{(b^{frac{1}{6}}+8)(b^{frac{1}{6}}-7)}{b^{frac{1}{6}}+7}=frac{b^{frac{1}{3}}+b^{frac{1}{6}}-56}{b^{frac{1}{6}}+7}}$