My question is really simple. We know that the definition of $\sqrt{a}$ is the positive solution of the equation $x^2=a$. What about $\sqrt[n]{a}$ for the others $n$?
Can I say if $n$ is odd than $\sqrt[n]{a}$ is the real solution of the equation $x^n-a=0$ and if $n$ is positive then $\sqrt[n]{a}$ is the positive real solution of the equation $x^n-a=0$?
So in this case, if I'm right, we will have $\sqrt[3]{-8}=-2$ and $\sqrt[4]{16}=2$
Yes, that's the standard definition of $\sqrt[n]a$.