I don't understand what they mean by "for negative values, $x^3=-\sqrt{x^6}$" (Khan Academy) I'm currently working on completing their first unit on calculus ab and I've encountered this roadblock. That's probably an exaggeration but I honestly can't figure out what they mean by "for negative numbers". I did the math and got the right number (at least the right absolute value) but the missing negative sign cost me the question and fair enough but why is there a negative sign that's being added anyway?$$\sqrt{x^6}=(x^6)^{1/2}=x^{6\times\frac12}=x^3$$I get that so why the negative?
for context here's their explanation and the problem itself
 A: 
I honestly can't figure out what they mean by "for negative numbers".

It means that the equation holds for negative $x$, that is if $x<0$.    Reason is that the real square-root is non-negative by definition:
$$\sqrt{x^2} = |x|$$
Now if $x < 0$, then $|x|= -x$ and thus $\sqrt{x^2} = -x= |x|$.  This still holds when we replace $x$ by $x^3$ due to $x<0 \iff x^3< 0$ and therefore:
$$\sqrt{x^6} = -x^3$$
And BTW it also holds for $x=0$.
A: The issue here has to due with non-integer exponents being weird. When we say $\sqrt{x}=y$, what we mean is that $y$ is a number which, when squared, gives $x$. However, there are two such numbers (unless $x=0$ of course). Both $y$ and $-y$ give $x$ when we square them. For example, $2^2 = (-2)^2=4$. We take the positive root by convention, but as far as exponentiation is concerned, there's not a natural reason to prefer it.
When we're dealing with non-integer exponents, the identity $$
(x^a)^b = x^{a b}
$$
really means that some value we could assign to the RHS is the same as some value we could assign to the LHS. It doesn't mean that the convention of taking positive roots necessarily makes the equation true. For example:$$
-1 = (-1)^{\frac12 \cdot 2} \ne ((-1)^2)^{\frac12} = \sqrt{1} = 1
$$
The same sort of thing is happening in this example. If we take $x<0$, then $x^3<0$, but $\sqrt{x^6} > 0$. So our convention failed us, and we need to take the other possible square root, i.e. $-\sqrt{x^6}$.
