Question

If the zeroes of the polynomial $p\left(x\right)=a{x}^3+3b{x}^2+3cx+d$ are in AP, then $2b^3-3abc+a^{2}d$ is equal to

• A. 3
• B. -3
• C. 0.0
• D. Can't be determined

Answer 1

The correct option is C 0.0

Let the zeroes be $\alpha ,\beta and\gamma$.

$\alpha +\beta +\gamma =\frac{-3b}{a}---\left(i\right)\alpha \beta +\beta \gamma +\gamma \alpha =\frac{3c}{a}---\left(ii\right)\alpha \beta \gamma =\frac{-d}{a}---\left(iii\right)$

Now, $\alpha ,\beta and\gamma$ are in AP.
$\Rightarrow 2\beta =\alpha +\gamma$

Substituting in Eq. $\left(i\right)$, $3\beta =\frac{-3b}{a}\Rightarrow \beta =\frac{-b}{a}$

Substituting in Eq. $\left(iii\right)$,
$\alpha \beta \gamma =\frac{-d}{a}\Rightarrow \left(\alpha \gamma \right)\left(\frac{-b}{a}\right)=\frac{-d}{a}\Rightarrow \alpha \gamma =\frac{d}{b}$

Substituting in Eq. $\left(ii\right)$,
$\beta (\alpha +\gamma )+\alpha \gamma =\frac{3c}{a}\Rightarrow \frac{-b}{a}\times (2\times \frac{-b}{a})+\frac{d}{b}=\frac{3c}{a}[\because \alpha +\gamma =2\beta =2\times \frac{-b}{a}]\Rightarrow 2\frac{b^2}{a^2}+\frac{d}{b}=\frac{3c}{a}\Rightarrow 2b^3+d{a}^2=3abc\Rightarrow 2b^3-3abc+a^{2}d=0$