Polynomial with geometric progression

Question

If the roots of an equation with real coefficients $ax^3+bx^2+cx+d=0, a\ne0$ form a geometric progression, prove $ac^{3}=db^{3}$

I have no idea how to get started, so if it can a few hints?

Answer 1

Start here:

$$(k_1 - x) (k_2 k_1 - x) (k_2^2 k_1 - x) = a x^3 + b x^2 + c x + d = 0$$

Answer 2

Say, $x_1,x_2,x_3$ are the roots, then by Vieta we have $$x_2^3= x_1x_2x_3 = -{d\over a}$$ and $$x_2(x_1+x_2+x_3) ={c\over a}$$ and $$x_1+x_2+x_3 =-{b\over a}$$ So from the alst two equation we get $x_2=-{c\over b}$ so putting this one in first equation we get $$-{b^3\over c^3} = -{d\over a}$$ and thus a conclusion.