It seems when a value in a radical is positive it's valid to "move" the exponent out of the radical.
Consider the function $\sqrt{x^2}$. When $x\geq0$ then
$\sqrt{x^2} = (\sqrt{x})^2$
For example, when $x = 5$
$\sqrt{5^2} = (\sqrt{5})^2$
$\sqrt{25} = \sqrt{5} \cdot \sqrt{5}$
$5 = 5$
However, when $x<0$ it's no longer valid to "move" the exponent out.
For example, when $x=-5$
$\sqrt{(-5)^2} \not= (\sqrt{-5})^2$
$\sqrt{25} \not= \sqrt{-5} \cdot \sqrt{-5}$
$5 \not= -5$
Also, it seems if a value inside a radical can be rendered positive, then an exponent can be "moved" out.
For example,
$\sqrt{(x^2)^3} = (\sqrt{x^2})^3 = |x|^3$
Thus, is it correct to say that if a value inside a radical is positive or can be rendered positive, an exponent can be "moved" out from a radical?
I'm guessing if this is the case, then it has something to do with the product rule of radicals which requires the values inside the radicals to be positive beofre being "combined":
if $a \ge 0$ and $b \ge 0$, then $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$
Best Answer
Yes, it is correct, and it follows from the properties of powers:
$$(x^{\alpha})^{\beta}=(x^{\beta})^{\alpha}$$
whenever all the quantities involved are well defined, e.g. for positive $x$ and real $\alpha,\beta$.