Extracting root of $y^3=-x^2$ Given $y^3=-x^2$ I thought the solution was $y=-x^{(2/3)}$ but as I see from the solution of my prof He wrote $y=-|x|^{(2/3)}$.
I don’t understand the absolute value since we’re extracting an odd root, what am I missing?
 A: Both expressions are actually equivalent. Note:
$y = −x^{(2/3)}$
$\implies y = -(\sqrt[3]{x})^2$
Taking the absolute value of x doesn't change the function as we square the term anyway.
A: if $x$ is a solution, the $|x|$ is a solution as well since
$|x|^{2/3}= (|x|^2)^{1/3} = (x^2)^{1/3}=x^{2/3}$
A: You are correct.
$$y = (-x^2)^{1/3} = (-1)^{1/3}(x^2)^{1/3}=-(x^2)^{1/3}=-x^{2/3}$$
The absolute value was brought in to make sure what was already sure, not really necessary.
A: I had a similar confusion recently.
See:
How exactly is the function $x^a$ defined?
So the key thing to realize here is this:

NOTE: In real analysis when they use/write $x^a$ (for any $a$ real) they
usually define it just for $x \ge 0$. This is to avoid all sorts of complications and ambiguities.

And when I say for any $a$ real I mean the power $a$ can be positive, negative, zero, rational or irrational.
So the way you have written your answer it implies $x \ge 0$ which is not necessarily true.
And it's in that sense (as others said) that the professor is more correct.
Important here is to realize that the NOTE kicks in when you're in real analysis.
