The condition that the roots of $x^3-b x^2+c x-d=0$ are in arithmetic progression is
The condition that the roots of $x^3-b x^2+c x-d=0$ are in arithmetic progression is
$9 c b=2 b^3+27 d$
$9 c b=2 d^3+27 b$
$9 c d=2 d^3+27 b$
$9 c d=2 b^3+27 d$
Solution
Let roots in A.P. be as $\alpha-r, \alpha, \alpha+r$
Sum of roots $=b$
$\Rightarrow \alpha-r+\alpha+\alpha+r=b \Rightarrow \alpha=\frac{b}{3}$
$\alpha$ is root of $x^3-b x^2+c x-d=0$
$\Rightarrow \frac{b^3}{27}-\frac{b^3}{9}+\frac{b c}{3}-d=0 \Rightarrow 9 c b=2 b^3+27 d$