The value of the cubic root of $-i$ So this was the question given to us. $\left(\iota=\sqrt{-1}\right)$ > Value(s) of $\left(-\iota\right)^{\dfrac{1}{3}}$ are > > (A) $\dfrac{\sqrt{3}-\iota}{2}$ > > (B) $\dfrac{\sqrt{3}+\iota}{2}$ > > (C) $\dfrac{-\sqrt{3}-\iota}{4}$ > > (D) $\dfrac{-\sqrt{3}+\iota}{2}$ And the answer, we were told, is option (A). I agree with this option; it can be easily obtained by putting $\left(-\iota\right)$ in polar form. But, my question is, why can't we rewrite $\left(-\iota\right)$ as $\iota^{3}$, and hence say that $\left(-\iota\right)^{\dfrac{1}{3}} = \iota $ ?

You are right: $i$ is another solution of the equation $z^3=-i$. There is also a third one, $\dfrac{1+i\sqrt{3}}{2}$.

However, this being a multiple choice question, your choices are constrained to the options given. Only one of the three roots was among the options.