Question

Prove that:
$(\sqrt{3\times 5^{-3}}÷\sqrt[3]{33^{-1}}\sqrt{5})\times \sqrt[6]{63\times 5^6}=\frac{3}{5}$

Answer 1

Given,

$\frac{\sqrt{3\times 5^{-3}}}{{\sqrt[3]{33}}^{-1}\sqrt{5}}\times \sqrt[6]{63\times 5^6}$

$=\frac{3^{\frac{1}{2}}\times 5^{\frac{-3}{2}}}{3^{\frac{-1}{3}}\times 5^{\frac{1}{2}}}\times 3^{\frac{1}{6}}\times 5^{\frac{6}{6}}[\because \sqrt{a}=a^{\frac{1}{2}}\text{and}(a^m{)}^n=a^{m\times n}]$

$=3^{\frac{5}{6}}\times 5^{-2}\times 3^{\frac{1}{6}}\times 5^1[\because \frac{a^m}{a^n}=a^{m-n},a^m\times a^n=a^{m+n}]$

$=3\times 5^{-1}$

$=\frac{3}{5}[\text{Using the identity}{a}^{-1}=\frac{1}{a}]$

$\frac{\sqrt{3\times 5^{-3}}}{{\sqrt[3]{33}}^{-1}\sqrt{5}}\times \sqrt[6]{63\times 5^6}=\frac{3}{5}$