For the equation $3x^2+px+3=0,p>0$, if one of the roots is square of the other, then $p$ is equal to

Let $\alpha ,{\alpha }^2$ be the roots of $3x^2+px+3$

$\therefore$ Product of the roots, 

$\alpha ·{\alpha }^2=1\Rightarrow {\alpha }^3=1$

Again, $\alpha +{\alpha }^2=-\frac{p}{3}$

$\Rightarrow 1+1=-\frac{p}{3}$   (if $\alpha =1$)

$\Rightarrow p=-6$

But $p>0$

$\therefore \alpha =1$ is not possible.

If $\alpha =\omega$, then $\alpha +{\alpha }^2=\omega +{\omega }^2=-1$

$\therefore -1=-\frac{p}{3}\Rightarrow p=3$

Again, if $\alpha ={\omega }^2$, then

$\alpha +{\alpha }^2={\omega }^2+{\omega }^4={\omega }^2+\omega =-1$

$\therefore -1=-\frac{p}{3}$

$\Rightarrow p=3$