If one root of $x^2+px+1=0$  is the cube of the other root, then $p=$

Let the roots are ${\alpha }^3,\alpha$

 then $\alpha +{\alpha }^3=-p$

${\alpha }^4=1$

${\alpha }^4-1=0$

$\left({\alpha }^2+1\right)\left({\alpha }^2-1\right)=0$

 then ${\alpha }^2=-1\left(or\right)\alpha =\pm 1$

If $\alpha =\pm 1$

 then $p=\pm 2$

If ${\alpha }^2=-1$

$\Rightarrow \alpha =\pm i$ $\left(\because -1=i^2\right)$

 then $p=0$