How to find $\sqrt[3]{8i}$ How do I find the following cube root?
$$\sqrt[3]{8i} = ?$$
I tried by adding $\sqrt[3]{i^3 + 8i + i}$ but that is where my imagination quits.

Here is another way that avoids exponential functions 
for this problem but is not a method that I would 
recommend in general, e.g. to find $\sqrt[10]{8i}$,
for which the method described Andre Nicolas's and
Agusti Roig's answers work much better.
Suppose $a$ and $b$ are real numbers such that $(a+bi)^3 = 8i$.  We have
$$\begin{align*}
(a+bi)^3 &= a^3 + 3a^2bi + 3a(bi)^2 + (bi)^3\\
&= (a^3 -3ab^2) + (3a^2b - b^3)i\\
&= 0 + 8i,
\end{align*}$$
and so
$$\begin{align*}
a^3 - 3ab^2 = a(a^2 - 3b^2) &= 0\\
3a^2b - b^3 = b(3a^2 - b^2) &= 8
\end{align*}$$


*

*From the first equation we see that one possible value for $a$ is $0$,
and in this case, the second equation reduces to $-b^3 = 8$ giving $b = -2$.
In other words, one cube root of $8i$ is $-2i$.

*Alternatively, if $a \neq 0$, then from the first equation we see 
that it must be that  $a^2 - 3b^2 = 0$
so that $a = \pm \sqrt{3}b$.  Substituting into the second equation, we 
get  $b(9b^2 - b^2) = 8b^3 = 8$, that is $b^3 = 1$. Now, $b = 1$ is 
the only real number solution to $b^3 = 1$, and so we get that 
the other two
cube roots of $8i$ are $\sqrt{3} + i$ and $-\sqrt{3} + i$. 
