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

By the Vieta's formulas we have $-\frac{b}{a}=(\alpha-\beta)+\alpha+(\alpha+\beta)=3\alpha$, $\frac{c}{a}=(\alpha-\beta)\alpha+(\alpha-\beta)(\alpha+\beta)+\alpha(\alpha+\beta)=$

$=\alpha^2-\alpha\beta+\alpha^2-\beta^2+\alpha^2+\alpha\beta=3\alpha^2-\beta^2$, $-\frac{d}{a}=(\alpha-\beta)\alpha(\alpha+\beta)=\alpha^3-\alpha\beta^2$

Then $(2b^2-9ac)b+27a^2d=a^3\left(2\left(\frac{b}{a}\right)^2-9\frac{c}{a}\right)\frac{b}{a}+27a^3\frac{d}{a}$

So we need to prove that $\left(2\left(\frac{b}{a}\right)^2-9\frac{c}{a}\right)\frac{b}{a}+27\frac{d}{a}=0$

We have $2\left(\frac{b}{a}\right)^2-9\frac{c}{a}=2(3\alpha)^2-9(3\alpha^2-\beta^2)=-9\alpha^2+9\beta^2$, so $\left(2\left(\frac{b}{a}\right)^2-9\frac{c}{a}\right)\frac{b}{a}+27\frac{d}{a}=$

$=(-9\alpha^2+9\beta^2)(-3\alpha)+27(-\alpha^3+\alpha\beta^2)=0$