Is the cube root of $a^3$ always a My question is as follows: for any real number a, is the cube root of the cube of a always equal to a?
For a = 2, the result seems to hold. However for a = -1, wolfram alpha says the cube root of -1 is non real? But,  $-1 = -1\times-1\times-1$ right? So surely the cube root of -1 is -1 and then shouldn't this also fit the statement: is the cube root of the cube of a always equal to a?
Perhaps I am confusing myself somewhere and there is something silly I am missing.
Thanks :)
 A: There are different possible "cube root" functions. They give the same result when the input is on the positive real axis, but differ elsewhere.
As long as we're considering only real inputs, the natural cube root function is simply the inverse function of $x\in\mathbb R \mapsto  x^3$. When we're using that, it is true by definition that $\sqrt[3]{a^3}=a$ for all real $a$.
However, Wolfram Alpha is not smart enough to notice that you probably want the real cube root, so it uses a different function that gives complex outputs for negative real inputs, but on the other hand is continuous on in the complex plane away from the real axis.
You will notice that the result it gives you indeed is a cube root of $-1$: $\left(\frac 12 + \frac{\sqrt 3}{2}i\right)^3 = -1$
A: The cube root of -1 is certainly -1.
For real numbers, you can only take the square root of a number greater than or equal to 0, but you can take the cube root of any number.
A: Over the real numbers, each number has exactly one cube root and the cubed root of a number cubed is itself. But over the complex numbers, each real number has exactly 3 cube roots and only one of those is real. Wolfram Alpha is probably considering the general case of complex numbers.
A: The fundamental theorem of algebra states that a polynomial of degree $n$ has at least $n$ roots up to multiplicity over an algebraically closed field.
In simpler terms, it means that the polynomial $a^3 = -1$ has at least three roots.
One of these roots is $-1$: $(-1)^3 = -1$.
However, it also has two other roots that are complex; specifically, the rotation of $-1$ about the origin in the complex plane by $\frac{2\pi}{3}$ radians, twice. So that's how you get the other two roots.
For if $z = \left[\exp i\frac{5\pi}{3}\right]^3$, then $z^3 = \exp i\frac{3\cdot 5\pi}{3} = \exp 5i\pi = \exp i\pi = -1$, and so on.
