Given equation is
$x^{3}+a x^{2}+b x+c=0$
Let $(\alpha, \beta, \gamma)$ be the roots of the given equation then according to given condition, we have
$2 \beta=\alpha+\gamma\,\,\,\,\,\,\,\,\,...(i)$
$(\because \alpha, \beta, \gamma$ are in $AP)$
Now, $\alpha+\beta+\gamma=-a$
$\Rightarrow \beta+2 \beta=-a\,\,\,\,$(Using Eq.(i)]
$\Rightarrow \beta=-\frac{a}{3}$
Since, $\beta$ is a root of given equation.
So, $\beta^{3}+a \beta^{2}+b \beta+c=0$
$\Rightarrow \left(-\frac{a}{3}\right)^{3}+a\left(\frac{-a}{3}\right)^{2}+b\left(\frac{-a}{3}\right)+c=0 $
$\Rightarrow -\frac{a^{3}}{27}+\frac{a^{3}}{9}-\frac{a b}{3}+c=0 $
$\Rightarrow -a^{3}+3 a^{3}-9 \,a b+27 c=0 $
$\Rightarrow 2 a^{3}-9 \,a b+27\, c=0 $
$\therefore 2 a^{3}-9 \,a b=-27 c$