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

The given equation is $x^2+px+1=0$

Given that one root is cube of the other root. 

suppose that the roots are $\alpha ,{\alpha }^3$

Sum of the roots is $\alpha +{\alpha }^3=-p$ and the product of the roots is $\alpha ·{\alpha }^3=1\Rightarrow {\alpha }^4=1$

Hence, $\alpha =1 ,-1 i,-i$

When $\alpha =1$, $p=-2$

When $\alpha =-1$, $p=2$

When $\alpha =\pm i$, the value of $p$is zero.

Therefore, the possible value of $p$ are $\boxed{\mathbf{0}\mathbf{,}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{2}}$