Question

4. 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

$az^3+bz^2+cz+d=0$

$z_1=\alpha -\beta, \ z_2=\alpha , \ z_3=\alpha+\beta$

Vieta's formulas:

$\begin{cases} z_1+z_2+z_3=-\frac{b}{a} \\ z_1 z_2+z_2 z_3+z_3 z_1=\frac{c}{a} \\ z_1z_2z_3=-\frac{d}{a} \end{cases}$

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

$\begin{cases} b=-3\alpha a \\ c=(3\alpha^2-\beta^2)a \\ d=-(\alpha^3-\alpha \beta^2)a \end{cases}$

$(2b^2-9ac)b+27a^2d=(2(-3\alpha a)^2-9a(3\alpha^2-\beta^2)a)(-3\alpha a)-27a^2(\alpha^3-\alpha \beta^2)a =(18\alpha ^2a^2-27\alpha^2a^2+9\beta ^2 a^2)(-3\alpha a)-27\alpha^3a^3+27\alpha \beta^2 a^3=27\alpha^3a^3 -27\alpha \beta^2a^3 -27\alpha^3a^3+27\alpha \beta^2 a^3=0$