# How to compute $(-1)^{1/3}$? [duplicate]

$(-1)^{1/3} = ((-1)^2)^{2/6} = 1^{2/6} = 1$

but the actual answer is (-1)

what is wrong with the first approach?

• There are several problems: $2\cdot 2/6 \neq 1/3$. Also $(a^b)^c = a^{bc}$ is in general not true. – mrf Mar 22 '16 at 12:24
• If you square and then take the (positive) square root you don't end up with the original number if it was negative. – almagest Mar 22 '16 at 12:25
• In my classes, I learned that it is unappropriate to write the $1/3$ power of a negative number, while $\sqrt[\frac{1}{3}]{-1}$ is correct – Laurent Duval Mar 22 '16 at 12:33

$$-1 = e^{i\pi}$$

Then in the general case

$$(-1)^n = \left(e^{i\pi}\right)^n = e^{i n\pi}$$

What could you say then?

Anyway

You're doing the cubic root of a negative number, hence the answer will be a negative number.

$$(-1)\cdot(-1)\cdot(-1) = -1$$

hence

$$(-1)^{1/3} = \sqrt[3]{-1} = -1$$

Consider the sixth roots of unity. Specifically the first, the third, and the fifth, sixth roots of unity.

Raising anyone of these numbers to the third power will get you to $-1$. Thus, any one of them is a cube root of $-1$.

In general, exponent rules do not work when the base is negative. In fact, if you check carefully, whenever exponent rules are discussed, the base is always assumed to be greater than zero.