Question

If the zeros of the polynomial $f\left(x\right)=a{x}^3+3b{x}^2+3cx+d$ are in A.P., then prove that $2b^3-3abc+a^{2}d=0$.

Answer 1

Let the zeros of the given polynomial be $p,q,r$. As the roots are in A.P., then it can be assumed as $p-k,p,p+k$, where $k$ is the common difference.

$p-k+p+p+k=-\frac{3b}{a}$

$p=-\frac{b}{a}$

And, $\left(p-k\right)\left(p\right)\left(p+k\right)=-\frac{d}{a}$
$p\left(p^2-k^2\right)=-\frac{d}{a}$

$-\frac{b}{a}\left(p^2-k^2\right)=-\frac{d}{a}$

$p^2-k^2=\frac{d}{b}$

And, $p\left(p-k\right)+p\left(p+k\right)+\left(p-k\right)\left(p+k\right)=\frac{3c}{a}$

$2p^2+\left(p^2-k^2\right)=\frac{3c}{a}$

$\frac{2b^2}{a^2}+\frac{d}{b}=\frac{3c}{a}$

$\frac{2b^3+a^{2}d}{a^{2}b}=\frac{3c}{a}$

$2b^3-3abc+a^{2}d=0$