# How do you simplify -8 ^ (2/3) ?

Aug 11, 2016

$- 4$

#### Explanation:

Using the $\textcolor{b l u e}{\text{laws of exponents}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{m} \times {a}^{n} = {a}^{m + n}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(A\right)$

Let's define, generally, the meaning of ${a}^{\frac{2}{3}}$

Using (A)

${a}^{\frac{2}{3}} \times {a}^{\frac{2}{3}} \times {a}^{\frac{2}{3}} = {a}^{\frac{6}{3}} = {a}^{2.} \ldots \ldots . \left(1\right)$

now ${a}^{\frac{2}{3}} \times {a}^{\frac{2}{3}} \times {a}^{\frac{2}{3}} = {\left({a}^{\frac{2}{3}}\right)}^{3.} \ldots \ldots . \left(2\right)$

Since the 2 expressions are equivalent we can equate them.

$\Rightarrow {\left({a}^{\frac{2}{3}}\right)}^{3} = {a}^{2}$

Take the $\textcolor{b l u e}{\text{cube root}}$ of both sides

$\Rightarrow \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{\frac{2}{3}} = \sqrt[3]{{a}^{2}} = {\left(\sqrt[3]{a}\right)}^{2}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

We can now evaluate $- {8}^{\frac{2}{3}}$

$- {8}^{\frac{2}{3}} = - 1 \times {\left(\sqrt[3]{8}\right)}^{2} = - 1 \times {\left(2\right)}^{2} = - 1 \times 4 = - 4$