<s>[INST] Prove that if the roots of $x^3+ax^2+bx+c=0$ form an arithmetic sequence, then $2a^3+27c=9ab$ 
Prove that if the roots of $x^3+ax^2+bx+c=0$ form an arithmetic sequence, then $$2a^3+27c=9ab$$

So far, I let the roots of $x^3+ax^2+bx+c=0$ be $r_1, r_2,$ and $r_3$. $r_1=r_2-d$ and $r_3=r_2+d$ because they form an arithmetic sequence with $d$ being the difference. the sum of the roots is $-a$. So, $r_2=-a/3$. We can let the product of the roots be $-c$. So, $(r_2-d)(r_2)(r_2+d)=-c$. Plugging in $r_2=-a/3$ we get $(-a/3-d)(-a/3)(-a/3+d)$. How do I continue with this method?
EDIT: I used hamam_abdallah's hint to get $\frac{-a^3}{27} + \frac{ad^2}{3} = -c$ what do i do after applying vieta's formulas?
 [/INST] Put $$a=3A$$ The equality to prove $$2a^3+27c=9ab$$ becomes
$$\boxed{2A^3+c=Ab}$$
As you said $r_2=\frac{-a}{3}= -A$ is a root of
$$x^3+3Ax^2+bx+c=0$$  then
$$(-A)^3+3A(-A)^2+b(-A)+c=0$$
$$\iff \; -A^3+3A^3+c=Ab$$
$$\iff \; \boxed{2A^3+c=Ab}$$
Done.
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