Where is -6?

Rewrite the equation as $3\left( x-{ x }_{ 1 } \right) \left( x-{ x }_{ 2 } \right)$ . Expanding it we see that $p=-3\left( { x }_{ 1 }+{ x }_{ 2 } \right)$ .

We also know that ${ x }_{ 1 }={ x }_{ 2 }^{ 2 }$ . Using Vieta's formulas ${ x }_{ 1 }\cdot { x }_{ 2 }=1\Rightarrow { x }_{ 1 }^{ 3 }=1\Rightarrow \left( { x }_{ 1 }-1 \right) \left( { x }_{ 1 }^{ 2 }+{ x }_{ 1 }+1 \right) =0$ .

If ${ x }_{ 1 }=1$ then $p=-6$ so we have that ${ x }_{ 1 }^{ 2 }+{ x }_{ 1 }+1=0$ .

Again using Vieta's formulas we see that the roots add up to $-1$ so $p= \boxed{3}$ .

THE other way is to work with the roots finally we get a cubic in p variable...Thats an easy method..

nice approach !!!!! but can be solved in another way