a degree 3 polynomial has two equal roots, prove..

p7 q13

question: a, b, c are real no. prove that if the equation

$\displaystyle x^3+ax^2 +bx+c= 0$

has two equal roots, then $\displaystyle 4(b^2-3ac)(a^2-3b)= (9c-ab)^2$

i know my method is not right:

let $\displaystyle f(x) = x^3+ax^2 +bx+c$

= $\displaystyle x^3+(2p+q)x^2+(p^2+2pq)x+p^2q$

so, $\displaystyle LHS= 4(b^2-3ac)(a^2-3b) = 4((p^2+2pq)^2-3(2p+q)p^2 q)((2p+q)^2 - 3(p^2+2pq)) =$ ......................

RHS = ...............

so finally LHS=RHS

what is the correct method to do it? thanks.