19 вар, 1,2,3 задания пожалуйста помогите упростить выражения

Question 1

Question 2

$( \frac{ (\sqrt[3]{x}-\sqrt[3]{a})^3+2x+a}{ (\sqrt[3]{x}-\sqrt[3]{a})^3-(x+2a)})^3+\sqrt{(a^3-3a^2x+3ax^2-x^3)^{ \frac{2}{3} }}:a=\frac{a-2x}{a}\\\\$

$\boxed{1} ( \frac{ (\sqrt[3]{x}-\sqrt[3]{a})^3+2x+a}{ (\sqrt[3]{x}-\sqrt[3]{a})^3-(x+2a)})^3=( \frac{ (\sqrt[3]{x})^3-3( \sqrt[3]{x})^2\sqrt[3]{a}+3\sqrt[3]{x}(\sqrt[3]{a})^2+2x+a }{(\sqrt[3]{x})^2\sqrt[3]{a}+3\sqrt[3]{x}(\sqrt[3]{a})^2-x-2a} )^3=\\\\ =(\frac{3a^{ \frac{2}{3}}-3\sqrt[3]{a}x^{ \frac{2}{3}}+3x}{3a^{ \frac{2}{3}}-3\sqrt[3]{a}x^{ \frac{2}{3}}-3a})^3=(\frac{3\sqrt[3]{x}(a^{ \frac{2}{3} }-\sqrt[3]{a}\sqrt[3]{x}+x^{ \frac{2}{3}})}{-3\sqrt[3]{a}(a^{ \frac{2}{3} }-\sqrt[3]{a}\sqrt[3]{x}+x^{\frac{2}{3}})})^3=(-\frac{\sqrt[3]x}{\sqrt[3]{a}})^3 =- \frac{x}{a}$

$\boxed {2} \sqrt{(a^3-3a^2x+3ax^2-x^3)^{ \frac{2}{3} }}:a= \sqrt{(a-x)^{3\cdot \frac{2}{3}}}:a= \sqrt{(a-x)^2}:a=\frac{a-x}{a}$

$\boxed{3} \ \frac{a-x}{a} - \frac{x}{a}= \frac{a-2x}{a}$

*************************************

$( \sqrt{ \sqrt{m}- \sqrt{ \frac{m^2-9}{m}}}+\sqrt{ \sqrt{m}+\sqrt{ \frac{m^2-9}{m}}} )^2\cdot \sqrt[4]{\frac{m^2}{4}}, \ npu \ m=4$

$( \sqrt{ \sqrt{4}- \sqrt{ \frac{4^2-9}{4}}}+\sqrt{ \sqrt{4}+\sqrt{ \frac{4^2-9}{4}}} )^2\cdot \sqrt[4]{\frac{4^2}{4}}=\\\\ =( \sqrt{ 2- \sqrt{ \frac{7}{4}}}+\sqrt{ 2+\sqrt{ \frac{7}{4}}} )^2\cdot \sqrt{2}=\\\\ =(\sqrt{2- \frac{\sqrt7}{2}}+\sqrt{2+ \frac{\sqrt7}{2}})^2\cdot \sqrt{2}=( \frac{\sqrt{4-\sqrt7}+\sqrt{4+\sqrt7}}{\sqrt2} )^2\cdot \sqrt{2}=\\\\ =(\frac{(\sqrt{4-\sqrt7}+\sqrt{4+\sqrt7})(\sqrt{4-\sqrt7}+\sqrt{4-\sqrt7})}{\sqrt2(\sqrt{4-\sqrt7}-\sqrt{4+\sqrt7})})^2\cdot \sqrt{2}=\\\\ = (\frac{-2\sqrt7}{-2})^2\cdot \sqrt{2} =(\sqrt7)^2=7\sqrt{2}$