Evaluating $\left((-1)^{2/3}\right)^{3/2} $ two ways gives different results ($\pm1$)

Question

I came across the expression: $$\left((-1)^{2/3}\right)^{3/2} $$ If I use the power rule I get $$\left((-1)^{2/3} \right)^{3/2} = -1$$ If I first simplify as $$ (-1)^{2/3} = \left((-1)^{1/3}\right)^{2} = (-1)^{2}=1 $$ which further raised to the power $\frac{3}{2}$, the result equal to $1$.

Where is the mistake? If possible, explain within the realm of Real numbers.

My question is different from the suggested duplicate because I'm not alluding to complex numbers. If I'm doing it implicitly, I'm not aware of it.

To be precise, $(-1)^{ab}$ is not always equal to $((-1)^a)^b$. I believe this is a duplicate and has been discussed before. — Nothing special, Feb 07 '24 at 16:44
Does this answer your question? Why $\sqrt{-1 \cdot {-1}} \neq \sqrt{-1}^2$? (it is the first I have found, but there are plenty of question very similar to this) — Marco, Feb 07 '24 at 16:58

score 0 · Answer 1 · answered Feb 07 '24 at 16:44

0

With respect to the order of operation you have this (like below) $$\underbrace{(\underbrace{(-1)^ \frac{2}{3}}_{1})^ \frac{3}{2} }_{2}$$and it is $$(+1)^ \frac{3}{2}=1 $$

answered Feb 07 '24 at 16:44

Khosrotash

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Evaluating $\left((-1)^{2/3}\right)^{3/2} $ two ways gives different results ($\pm1$)

1 Answers1