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**NonCommAlg** let $\displaystyle x_0$ be the double root. then $\displaystyle f(x_0)=f'(x_0)=0,$ i.e. $\displaystyle x_0^3+ax_0^2 + bx_0+c=0,$ which we'll call it (1), and

$\displaystyle 3x_0^2 + 2ax_0+b=0,$ which we'll call it (2). if $\displaystyle x_0=0,$ then (1) and (2) gives us b = c = 0, and what you want

to prove obviously holds. so i'll assume that $\displaystyle x_0 \neq 0.$ multiply (1) by 3 and (2) by $\displaystyle -x_0$ and then add the result

together to get $\displaystyle ax_0^2+2bx_0+3c=0.$ call this (3). now multiply (2) by $\displaystyle a$ and (3) by $\displaystyle -3$ and then add the result

together to get $\displaystyle 2(a^2-3b)x_0-(9c-ab)=0.$ call this one (4). now multiply (2) by $\displaystyle 3c$ and (3) by $\displaystyle -b$ and then

add the result together to get $\displaystyle (9c-ab)x_0-2(b^2-3ac)=0.$ call this (5). finally multiply (4) by $\displaystyle 9c - ab$ and (5)

by $\displaystyle -2(a^2-3b)$ and add the result together to get $\displaystyle 4(b^2-3ac)(a^2-3b)=(9c-ab)^2. \ \ \ \square$