$O, A, B, C$ are the four points on straight line such that the distances of $A, B, C$ from $O$ are the roots of equation $ax^3+3bx^2+3cx+d=0$. If $B$ is the middle point of $AC$, show that $a^2d-3abc+2b^3=0$.
EDIT:
I solved it using Vieta's formula. But I am looking for an alternative solution
On $OABC$ straight line, I took $OA=q, AB=p \ and \ BC=p$.
Then the roots are $q, (p+q), (2p+q)$
$Σα = (\dfrac {-3b}{a})$
$Σαß = (\dfrac {3c}{a})$
$∏α = (\dfrac {-d}{a})$
$Σα= 3(p+q) = (\dfrac {-3b}{a}) \Rightarrow (p+q)= (\dfrac {-b}{a})$
$Σαß = q(p+q)+q(2p+q)+(p+q)(2p+q)=(\dfrac {3c}{a})$
$\Rightarrow 2(p+q)^2+q(q+2p)=(\dfrac {3c}{a})$
$\Rightarrow q(p+\overline{p+q})=\dfrac {3ac-2b^2}{a^2}$
$\Rightarrow q(p-\dfrac{b}{a})=\dfrac {3ac-2b^2}{a^2}$----(1)
$∏α = q(p+q)(2p+q)= (\dfrac {-d}{a})$
$\Rightarrow q(\dfrac {-b}{a})(p-\dfrac {b}{a})=\dfrac {-d}{a}$
$\Rightarrow q(p-\dfrac {b}{a})=\dfrac {d}{b}$------(2)
Equating (1) & (2)
$\dfrac {3ac-2b^2}{a^2}=\dfrac {d}{b}$
Solving which we get,
$a^2d-3abc+2b^3=0$
I am looking for alternative solutions.