Question

If the roots of the cubic az3 + bz2 + cz + d = 0 form an arithmetic progression α − β,
α, α + β, prove that (2b2 − 9ac)b + 27a2d = 0.

Answer 1

We are given that $az^3 + bz^2 + cz + d = 0$ has three roots in arithmetic progression $α − β, α, α + β$.

Now, we known as: Sum of roots = $-b/a$ , Product of roots = $-d/a$ and sum of product of two roots = $c/a$.

So, we got three equations in our problem as:

$(\alpha−\beta)+(\alpha)+(\alpha+\beta)= -\frac{b}{a} \\ (\alpha−\beta)(\alpha)(\alpha+\beta) = -\frac{d}{a} \\ (\alpha−\beta)(\alpha)+ (\alpha)(\alpha+\beta)+ (\alpha−\beta)(\alpha+\beta) = \frac{c}{a}$

So, $3\alpha = -\frac{b}{a}, \alpha(\alpha^2-\beta^2) = -\frac{d}{a}$ and $3\alpha^2-\beta^2 = \frac{c}{a}$ .

So by first and last equation we got, $\alpha = -\frac{b}{3a}$ and $\beta^2 = \frac{b^2}{3a^2}-\frac{c}{a}$.

Now from middle equation we get,

$(-\frac{b}{3a})(\frac{b^2}{9a^2} - \frac{b^2}{3a^2} + \frac{c}{a} ) = -\frac{d}{a}$

$\implies \frac{b(b^2 - 3b^2 + 9ac )}{9a^2} = 3d \\ \implies (2b^2 - 9ac)b + 27a^2d = 0$