Solve $z^5 +32 =0$ 
Solve $z^5 +32 =0$

My attempt : $$z^5 = -32$$
            Multiply the powers on both sides by $\frac{1}{5}$
we get $$z = 2 * (-1)^\frac{1}{5}$$
Now I'm stuck at this step I don't know how to proceed. Kindly help.

We have $$\left(-\dfrac z2\right)^5=1$$
Now $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$
If $x^5-1=0,$
either $x-1=0\iff x=1$
or $x^4+x^3+x^2+x+1=0$
Like  Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$,
as $x\ne0,$ divide both sides by $x^2$  to find $$0=x^2+\dfrac1{x^2}+x+\dfrac1x+1=\left(x+\dfrac1x\right)^2+\left(x+\dfrac1x\right)-1$$
Solve for $x+\dfrac1x$  then solve for $x$
