What are roots of higher power like $n^9=512$? We know a value of $n$ in $n^2 = 4$ is $\pm2$ (two roots), then what about $n^9 = 512$, generic answer is $+2$, but it should have $9$ roots. What are those $8$ other roots and how can we find them ?
 A: Since $1=e^{2k\pi i}$ for any integer $k$, you can write 
$$n = (512\cdot 1)^{1/9} = 512^{1/9}(e^{2k\pi i})^{1/9} = 2 e^{\frac{2k\pi i}{9}}, $$
for any integer $k$.  By Euler's formula this is
$$n = 2\left(\cos \frac{2k\pi i}{9} + i \sin \frac{2k\pi i}{9}\right) $$
and these complex numbers will all be different for $k= 0, 1, 2, \ldots, 8$,
giving you all 9 roots.
A: $2$ is not the only root here.The question is how to get all the roots in such a situation?? 
Here the concept of $nth$ roots of complex number work. Every number can be regarded as a complex number, If it is real just assume it to be a complex number having imaginary part$=0$
For the equation  $Z^n=z_0$ we have n roots which are:

$Z_k = n\sqrt{r} (\cos \frac{\theta +2k\pi}{n}+i\sin\frac{\theta +2k\pi}{n})$ where r is modulus of complex number.

Here are  all the roots of your equation.
A: factorizing $$n^9-512$$ we get $$\left( n-2 \right)  \left( {n}^{2}+2\,n+4 \right)  \left( {n}^{6}+8\,
{n}^{3}+64 \right) 
$$ and from here you will get all solutions.
A: $$n^9-a^9=(n^3)^3-(a^3)^3=(n^3-a^3)(n^6+a^3n^3+a^6)=(n-a)(n^2+an+a^2)(n^6+a^3n^3+a^6)$$
