<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] Hint
Finally note that
$$(-a/3-d)(-a/3+d) +(-a/3)(-a/3+d)+(-a/3-d)(-a/3)=b$$
gives
$$(-a/3-d)(-a/3+d)= b-(-a/3)(-a/3+d)-(-a/3-d)(-a/3)=b+ (a/3)(-a/3+d-a/3-d)$$
Thus
$$(-a/3-d)(-a/3+d)=b- (2a^2/9)$$
Therefore
$$-c=(-a/3-d)(-a/3)(-a/3+d)=(-a/3)\left( b- (2a^2/9) \right)$$
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