Question

If $x^3+px+q$ and ${3x}^2+p$ have a common factor then ${4p}^3+{27q}^2$ is equal to:-

Answer 1

$x^3+px+q$

$3x^2+p$ have common root

Let common root $=a$

$a^3+ap+q=0$

$a^2=\frac{-p}{3}$

$a=\sqrt{\frac{-p}{3}}$

$4p^3+27q^2=?$

$a^3(a^2+p)+q=0$

$\Rightarrow \sqrt{\frac{-p}{3}}[\left(\frac{-p}{3}\right)+p]+q=0$

$\sqrt{\frac{-p}{3}}\left(\frac{+2p}{3}\right)=-q$

$\Rightarrow \frac{-p\left(4p^2\right)}{27}=q^2$

$4p^3+27q^2=0$.