If the roots of the equation \(x^{3}+a x^{2}+b x+c=0\) are in \(\mathrm{AP}\), then \(2 a^{3}-9 a b\) is equal to

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
\(2 \beta=\alpha+\gamma...(i)(\because \alpha, \beta, \gamma are in AP)\)
\(\begin{array}{ll}
\text { Now, } & \alpha+\beta+\gamma=-a \\
\Rightarrow & \beta+2 \beta=-a (Using Eq. (i)\\
\Rightarrow & \beta=-\frac{a}{3}
\end{array}\)
Since, \(\beta\) is a root of given equation.
\(\begin{array}{lc}
\text { So, } & \beta^{3}+a \beta^{2}+b \beta+c=0 \\
\Rightarrow & \left(-\frac{a}{3}\right)^{3}+a\left(\frac{-a}{3}\right)^{2}+b\left(\frac{-a}{3}\right)+c=0 \\
\Rightarrow & -\frac{a^{3}}{27}+\frac{a^{3}}{9}-\frac{a b}{3}+c=0 \\
\Rightarrow & -a^{3}+3 a^{3}-9 a b+27 c=0 \\
\Rightarrow & 2 a^{3}-9 a b+27 c=0
\end{array}\)