Question

Simplify and express the answer with positive exponent.$\begin{array}{l}{[\sqrt[3]{x^{4}y}\times \frac{1}{\sqrt[3]{x{y}^7}}]}^{-4}\end{array}$

Answer 1

Simplification of algebric expression: $\begin{array}{l}{[\sqrt[3]{x^{4}y}\times \frac{1}{\sqrt[3]{x{y}^7}}]}^{-4}\end{array}$

$={\left[\left(x^{\frac{4}{3}}{y}^{\frac{1}{3}}\right).\left(\frac{1}{x^{{\displaystyle \frac{1}{3}}}{y}^{{\displaystyle \frac{7}{3}}}}\right)\right]}^{-4}$ $\left(\because \sqrt[n]{x}=x^{\frac{1}{n}}\right)$

$={\left[\left(x^{\frac{4}{3}}{y}^{\frac{1}{3}}\right).\left(x^{-\frac{1}{3}}{y}^{-\frac{7}{3}}\right)\right]}^{-4}$

$={\left[\left(x^{\frac{4}{3}}\times x^{-\frac{1}{3}}\right).\left(y^{\frac{1}{3}}\times y^{-\frac{7}{3}}\right)\right]}^{-4}$

$={\left[\left(x^{\frac{4}{3}-\frac{1}{3}}\right).\left(y^{\frac{1}{3}-\frac{7}{3}}\right)\right]}^{-4}$

$={\left[\left(x^{\frac{3}{3}}\right).\left(y^{-\frac{6}{3}}\right)\right]}^{-4}$

$={\left[\left(x\right).\left(y^{-2}\right)\right]}^{-4}$

$={\left[\frac{x}{y^2}\right]}^{-4}$

$={\left[\frac{y^2}{x}\right]}^4$

Therefore, $\begin{array}{l}{[\sqrt[3]{x^{4}y}\times \frac{1}{\sqrt[3]{x{y}^7}}]}^{-4}\end{array}$$={\left[\frac{y^2}{x}\right]}^4$.