# How do you write root3(x^-6) with fractional exponent?

Mar 20, 2016

${x}^{-} 2$

#### Explanation:

Note that

$\sqrt[a]{{x}^{b}} = {x}^{a / b}$

Thus,

$\sqrt{{x}^{-} 6} = {x}^{- 6 / 3} = {x}^{-} 2$

Note that

${x}^{-} 2 = \frac{1}{x} ^ 2$

both of which are acceptable answers.

Mar 20, 2016

Very slight expansion on the process

$\frac{1}{x} ^ 2 = {x}^{- 2}$

#### Explanation:

Given:$\text{ } \sqrt{{x}^{- 6}}$

Consider the ${x}^{- 6}$ this is the same as $\frac{1}{x} ^ 6$

Put it back into the root giving

$\text{ } \sqrt{\frac{1}{x} ^ 6}$

This is the same as

$\text{ } \frac{\sqrt{1}}{\sqrt{{x}^{6}}}$

But ${x}^{6} = {x}^{2 + 2 + 2} = {x}^{2} \times {x}^{2} \times {x}^{2} = {\left({x}^{2}\right)}^{3}$ giving

$\text{ } \frac{\sqrt{1}}{\sqrt{{\left({x}^{2}\right)}^{3}}} = \frac{1}{x} ^ 2$

$\text{'~~~~~~~~~~~~ Important Note ~~~~~~~~~~~~~~~~~~~~~~~}$

The key question is about the state being positive or negative.

Consider $\sqrt{1}$: for 1 to be negative there would need to be an indicator in the starting point that complex numbers are involved. This is not the case so this question has no credence.

Consider the ${x}^{2} \times {x}^{2} \times {x}^{2}$: if $x$ where to be negative then each ${x}^{2}$ would change it back to positive. So it is possible for $x < 1$, that is, the source value could be either negative or positive.

However the given expression has been simplified such that it involves ${x}^{2}$. This automatically results in an outcome that is positive.

So it is correct to write the answer as positive. That is, we do not have negative $\frac{1}{x} ^ 2$ 