Question

If the zeros of the polynomial $f\left(x\right)=x^3+3p{x}^2+3qr$ are in A.P, then the value of $2p^3+r$ for $p=q=1$ is $\frac{4}{b}$. So, $b$ is:

Answer 1

if $f\left(x\right)=x^3+3p{x}^2+3qr$ and its roots are in A.P., then roots are in the form $a-d,a,a+d$.

Sum of roots $=-3p$

Also, sum of roots$=(a-d)+\left(a\right)+(a+d)=3a$

$\Rightarrow 3a=-3p$

$\Rightarrow a=-p$

Now, we can write sum of roots two at a time as $(-p-d)(-p)+(-p)(-p+d)+(-p+d)(-p-d)=3p^2-d^2=0$

$3p^2=d^2$

Product of roots$=-3qr=(-p-d)(-p)(-p+d)=p{d}^2-p^3=p\left(3p^2\right)-p^3=2p^3$

As $p=q=1$,

$\Rightarrow -3r=2$ and $r=\frac{-2}{3}$

$\Rightarrow 2p^3+r=2(1{)}^3+\frac{-2}{3}=\frac{4}{3}$