# How do you simplify root5(-t^10 )?

Oct 6, 2015

$\sqrt[5]{- {t}^{10}} = - {t}^{2}$

#### Explanation:

If $a > 0$ and $b , c \ge 0$, then ${a}^{b c} = {\left({a}^{b}\right)}^{c}$

Also $\sqrt[n]{- a} = - \sqrt[n]{a}$ if $n \in \mathbb{Z}$ is odd.

So $\sqrt[5]{- {t}^{10}} = - \sqrt[5]{{t}^{10}} = - {\left({t}^{10}\right)}^{\frac{1}{5}} = - {t}^{10 \cdot \frac{1}{5}} = - {t}^{2}$

We can also see that ${\left(- {t}^{2}\right)}^{5} = {\left(- 1\right)}^{5} \cdot {\left({t}^{2}\right)}^{5} = - {t}^{10}$, so $- {t}^{2}$ is a fifth root of $- {t}^{10}$.