Решите пожалуйста с решением 10 класс

Решите задачу:

$\sqrt[3]{(-3)^3\cdot 2^6}=(-3)\cdot 2^2=-12\\\\\sqrt[4]{0,5}\cdot \sqrt[4]{0,125}=\sqrt[4]{0,5\cdot 0,125}=\sqrt[4]{(0,5)^4}=0,5\\\\\sqrt[3]{32}:2^{\frac{2}{3}}-\sqrt{121}=\frac{\sqrt[3]{2^5}}{2^{2/3}}-11=\frac{2^{5/3}}{2^{2/3}}-11=2-11=-9\\\\\frac{\sqrt[4]{y^3}}{(y^{\frac{1}{3}})^{\frac{9}{2}}}=\frac{y^{3/4}}{y^{3/2}}=\frac{1}{y^{3/4}}=\frac{1}{\sqrt[4]{y^3}}\\\\\frac{a^{1/2}-1}{a^{1/4}-1}-\sqrt[4]{a}=\frac{(a^{1/4}-1)(a^{1/4}+1)}{a^{1/4}-1}-a^{\frac{1}{4}}=a^{1/4}+1-a^{1/4}=1$

$\frac{y-16y^{\frac{1}{2}}}{5y^{\frac{1}{4}}+20}=\frac{y^{\frac{1}{2}}(y^{\frac{1}{2}}-16)}{5(y^{\frac{1}{4}}+4)}=\frac{y^{\frac{1}{2}}(y^{\frac{1}{4}}-4)(y^{\frac{1}{4}}+4)}{5(y^{\frac{1}{4}}+4)}=\frac{1}{5}\, y^{\frac{1}{2}}(y^{\frac{1}{4}}-4)=\\\\=\frac{1}{5} \sqrt{y}(\sqrt[4]{y}-4)$