Question

Answers please..for these 3 questions...
Guys please help me ​

Answer 1

(i)

$(\sqrt[3]{3}+\sqrt[3]{2})(3^{2/3}+2^{2/3}-6^{1/3})\\\\=(\sqrt[3]{3}+\sqrt[3]{2})(\sqrt[3]{3^2} +\sqrt[3]{2^2}-\sqrt[3]{6})\\\\=\sqrt[3]{3}*\sqrt[3]{3^2}+\sqrt[3]{3}*\sqrt[3]{2^2}-\sqrt[3]{3}*\sqrt[3]{6}+\sqrt[3]{2}*\sqrt[3]{3^2}+\sqrt[3]{2}*\sqrt[3]{2^2}-\sqrt[3]{2}*\sqrt[3]{6}\\\\=\sqrt[3]{3^3}+\sqrt[3]{12}-\sqrt[3]{18}+\sqrt[3]{18}+\sqrt[3]{2^3}-\sqrt[3]{12}\\\\=3+2\\\\=5$

(ii)

$\sqrt[5]{\sqrt[4]{(2^4)^3}}-5\sqrt[5]{8}+2\sqrt[5]{\sqrt[4]{(2^3)^4}}\\\\=\sqrt[5]{\sqrt[4]{(2^3)^4}}-5\sqrt[5]{2^3}+2\sqrt[5]{\sqrt[4]{(2^3)^4}}\\\\=3\sqrt[5]{\sqrt[4]{(2^3)^4}}-5\sqrt[5]{2^3}\\\\=3\sqrt[5]{(2^3)}-5\sqrt[5]{(2^3)}\\\\=2\sqrt[5]{(2^3)}\\\\=-2*2^{3/5}\\\\=-2^{1+3/5}$

(iii) incomplete ..... this is very hard for a middle school student lol ...

$\frac{243^r*9*(3^{-r/3})^2-(81)^r}{9^{2t}*2^3}=81^{-1}\\\\=>\frac{81^r*3^r*9*(3^{-r/3})^2-(81)^r}{9^{2t}*2^3}=81^{-1}\\\\=>\frac{81^r(3^r*9*(3^{-r/3})^2-1)}{81^{t}*2^3}=81^{-1}\\\\=>\frac{81^{r-t}(3^{r+2}*3^{-2r/3}-1)}{2^3}=81^{-1}\\\\=>81^{r-t}(3^{r+2-2r/3}-1)=81^{-1}*8$

answer not completed, sorry