$ x^3+3px^2+(4p-1)x+p=(x+1)(x-a)(x-b)$ Now it's easy

utkarshgupta wrote: $ x^3+3px^2+(4p-1)x+p=(x-1)(x-a)(x-b)$ Now it's easy I think you meant $x=-1$ (not $x=1$) to be a factor of the polynomial. Then, you have $x^3+3px^2+(4p-1)x+p=(x+1)(x^2+(3p-1)x+p)$. Now, either $0$ or $-2$ must be a root of the quadratic polynomial which yields easily $p=0$ or $p=\frac{6}{5}$ or the two roots of the quadratic must have difference 1, i. e. the discriminant has to be 1, hence $(3p-1)^2-4p=1$ which gives again $p=0$ or $p=\frac{10}{9}$.

http://www.komal.hu/verseny/feladat.cgi?a=feladat&f=A527&l=en

This was easy once you see that $-1$ is a root of the equation.

If you can't see the root readily (what if it was something complicated) Then I think observing, $p=\frac{x-x^3}{3x^2+4x+1}$ just might give you the hint Though I doubt if it would be of much help in tougher questions

Yeah, A good technique.

Just take the roots as a, a+1 and b. Equate to the coefficients and by adding them all we can factorise pretty easily. This is a pretty standard approach i guess.