Q. 258 - Intermediate Algebra

Question

In the following exercises, simplify.  

$\frac{\sqrt[3]{162x^{-3}{y}^6}}{\sqrt[3]{2x^3{y}^{-2}}}$

Step-by-Step Solution

Answer

The simplified form is $\frac{3y^2\sqrt[3]{3y^2}}{x^2}$.

Step 1. Given information.

The given expression is 

$\frac{\sqrt[3]{162x^{-3}{y}^6}}{\sqrt[3]{2x^3{y}^{-2}}}$.

Step 2. Simplifying.

Now, to simplify the expression first we will rewrite the expression using the quotient property.

$\frac{\sqrt[3]{162x^{-3}{y}^6}}{\sqrt[3]{2x^3{y}^{-2}}}=\sqrt[3]{\frac{162x^{-3}{y}^6}{2x^3{y}^{-2}}}$

Then removing the common factors.  

$\sqrt[3]{\frac{162x^{-3}{y}^6}{2x^3{y}^{-2}}}=\sqrt[3]{\frac{81y^8}{x^6}}$

On simplifying we get, 

$\sqrt[3]{\frac{81y^8}{x^6}}=\frac{3y^2\sqrt[3]{3y^2}}{x^2}$

Hence, the simplified form is $\frac{3y^2\sqrt[3]{3y^2}}{x^2}$