Q. 586. - Intermediate Algebra

Question

In the following exercises, simplify. Assume all variables are positive.

${\left(\frac{8x^{{\displaystyle \frac{2}{3}}}{y}^{-{\displaystyle \frac{5}{2}}}}{x^{-\frac{7}{3}}{y}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}$

Step-by-Step Solution

Answer

The simplified form of the given expression is $\frac{2x}{y}$.

Step 1. Given information.

The given expression is ${\left(\frac{8x^{{\displaystyle \frac{2}{3}}}{y}^{-{\displaystyle \frac{5}{2}}}}{x^{-\frac{7}{3}}{y}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}$

Step 2. Simplify the exponents.

Applying the Negative Exponent Property,

${\left(\frac{8x^{{\displaystyle \frac{2}{3}}}{y}^{-{\displaystyle \frac{5}{2}}}}{x^{-\frac{7}{3}}{y}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}={\left(\frac{8x^{{\displaystyle \frac{2}{3}}}{x}^{{\displaystyle \frac{7}{3}}}}{y^{\frac{5}{2}}{y}^{\frac{1}{2}}}\right)}^{\frac{1}{3}}$

Now we will use the Product property, 

$$\begin{gathered} {\left(\frac{8x^{{\displaystyle \frac{2}{3}+\frac{7}{3}}}}{y^{\frac{5}{2}+{\displaystyle \frac{1}{2}}}}\right)}^{\frac{1}{3}}={\left(\frac{8x^{{\displaystyle \frac{9}{3}}}}{y^{{\displaystyle \frac{6}{2}}}}\right)}^{\frac{1}{3}} \\ ={\left(\frac{8x^{{\displaystyle 3}}}{y^3}\right)}^{\frac{1}{3}} \end{gathered}$$

Let's use the Power Property,

$\left(\frac{8^{{\displaystyle \frac{1}{3}}}{x}^{3·{\displaystyle \frac{1}{3}}}}{y^{3·\frac{1}{3}}}\right)=\frac{2x}{y}$