Why isn't $\sqrt{64x^4y^8z^6}$ equal to $8x^2y^4z^3$?

Question

I simplified $$\sqrt{64x^4y^8z^6}$$ by taking the square root of $64$ (getting $8$), $x^4$ (getting $x^2$), $y^8$ (getting $y^4$), and $z^6$ (getting $z^3$). My answer, $8x^2y^4z^3$, not quite right and I am having a hard time understanding why not.

I even placed the problem into WolframAlpha.com (here) and got the same answer, which was really confusing. Any pointers of what I might be doing wrong?

Answer 1

As per the hint, you have to be careful since $z^3$ is negative if $z<0$ and the square root refers to the principal (nonnegative) root. Wolfram also mentions the assumption that $x,y,z$ are positive in their simplification to your answer. But for arbitrary real $x,y,z$ we have

$$\sqrt{64x^4y^8z^6}=8x^2y^4|z|^3.$$

Answer 2

It’s conventional for $\sqrt n$ to give you the positive square root of $n$, but in this case they might be expecting you to give the answer in the form $\pm {8x^2y^4z^3}$