Is $\sqrt[3]{\left(\frac{2}{3}\right)^{2}}$ equivalent to both $\left(-\frac{2}{3}\right)^{2/3}$ and $\left(\frac{3}{2}\right)^{-2/3}$?

Question

A problem on my midterms was multiple choice with 4 choices to choose from. The problem asked for what is equivalent to $$\sqrt[3]{\left(\frac{2}{3}\right)^{2}}$$ and there were two possible answers that were $$\left(-\frac{2}{3}\right)^{2/3}\quad \text{and}\quad \left(\frac{3}{2}\right)^{-2/3}$$ I claimed that these are equal.

So after the test I pulled up Desmos, plugged them in, and they all gave me 0.763142828369.

So then I asked wolfram alpha and I put in the -2/3^2/3 thing and got me an imaginary result. But I wasn't convinced because the cube root of any number is real and (-a)^2 = a^2. so i guess wolfram didn't do each step, and considered 2/3 as its own thing idk. but I do know that that's how sir taught us how to do fractional exponents, which is step by step and he would probably square -2/3 first and then do the rest.

But when I asked any teacher about the problem, they all said no with no explanation they didn't have any second thought. They were confident, so I decided I was confident with my answer.

So my real problem is either getting a more mature mathematician ('cause I'm only 16, and who would listen to what a 16 year old says) to defend me or I should just wait until schools open and I can discuss it with a teacher.

Answer 1

Recall that the cube root of $x$, denoted $\sqrt[3]{x}$ or $x^{1/3}$, is a number $z$ that satisfies the equation $z^3 = x$.

Note that I wrote “a number” instead of “the number”. Because, according to the Fundamental Theorem of Algebra, a polynomial of degree 3 can have as many as 3 distinct roots.

By convention, $\sqrt[3]{x}$ denotes the “principal” cube root of $x$. If you need the other two, they're $\omega\sqrt[3]{x}$ and $\overline{\omega}\sqrt[3]{x}$, where $\omega$ and $\overline{\omega}$ denote $-\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$, the two non-real cube roots of 1.

This raises the question: Which of the three cube root is “principal”?

When $x$ is a nonnegative real number, this is straightforward: Pick the one that's real.

Otherwise, let $r = |x|$ and $\theta = \arg(x)$, so that $x = re^{i\theta} = r(\cos\theta + i\sin\theta)$. Then, by DeMoivre's Formula,

$$\boxed{\sqrt[3]{x} = \sqrt[3]{r}(\cos\frac{\theta}{3} + i\sin\frac{\theta}{3})}$$

So, for example, because $x=i$ gives $r=1$ and $\theta=\frac{\pi}{2}$,

$$\sqrt[3]{i} = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6} = \frac{\sqrt{3}}{2} + i\frac{1}{2}$$

And for $x=-1$, $r = 1$ and $\theta = \pi$, so:

$$\sqrt[3]{-1} = \cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = \frac{1}{2} + i\frac{\sqrt{3}}{2}$$

But wait! That means that the cube root of a real number can be a non-real number. What if you don't want complex numbers in your calculations?

Well, one way is to special-case the definition of the principal cube root so that $\sqrt[3]{-x} = -\sqrt[3]{x}$. For example, this is what a TI-89 or TI-92 calculator does when the complex number mode is set to “real”.

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Now, let's consider what the expression $x^{2/3}$ means if $x$ is a negative real number.

If you calculate it as $(x^2)^{1/3}$, then the squaring cancels out the negation and you just get $|x|^{2/3}$.

But, if you use DeMoivre's or Euler's definition of complex exponentiation, representing the radicand in terms of $r$ and $\theta$, you get $r^{2/3}(\cos\frac{2\theta}{3} + i\sin\frac{2\theta}{3})$

And this is why Python and WolframAlpha give a non-real answer when you calculate $(-\frac{2}{3})^{2/3}$.

>>> (-2/3)**(2/3)
(-0.3815714141844438+0.6609010760833648j)

Point is, while an expression of the form $a^b$ is unambiguous when $a > 0$, things can get complex (pun intended) when $a < 0$.