Помогите упростить, никак не получается(

Используем свойства степеней $x^m \cdot x^n= x^{m+n}; \ \ \frac{x^m}{x^n}=x^{m-n}; \ \ (x^n)^m=x^{n \cdot m}$

$\frac{(b^{\frac{7}{5}}-b^{-\frac{3}{5}})\cdot b^{\frac{3}{5}}}{b^{\frac{1}{3}}\cdot (b^{\frac{2}{3}}+b^{-\frac{1}{3}})}=\frac{b^{\frac{7}{5}}-b^{-\frac{3}{5}}}{b^{\frac{2}{3}}+b^{-\frac{1}{3}}} \cdot b^{\frac{3}{5}-\frac{1}{3}}=\frac{b^{\frac{7}{5}}-b^{-\frac{3}{5}}}{b^{\frac{2}{3}}+b^{-\frac{1}{3}}} \cdot b^{\frac{4}{15}}=\frac{b^{(\frac{7}{5} + \frac{4}{15})}-b^{(-\frac{3}{5}+ \frac{4}{15})}}{b^{\frac{2}{3}}+b^{-\frac{1}{3}}}$ $\frac{b^{\frac{25}{15}}-b^{-\frac{5}{15}}}{b^{\frac{2}{3}}+b^{-\frac{1}{3}}} =\frac{b^{\frac{5}{3}}-b^{-\frac{1}{3}}}{b^{\frac{2}{3}}+b^{-\frac{1}{3}}} =\frac{b^{\frac{5}{3}}-\frac{1}{b^{\frac{1}{3}}}}{b^{\frac{2}{3}}+\frac{1}{b^{\frac{1}{3}}}}=\frac{b^2-1}{b+1}=\frac{(b-1)\cdot(b+1)}{b+1}=b-1$