Prove that the roots of $x^3+ax^2+bx+c=0$ form an arithmetic sequence iff $2a^3+27c=9ab$.

Question

1. Prove that if the roots of $x^3 + ax^2 + bx + c = 0$ form an arithmetic sequence, then $2a^3 + 27c = 9ab$.

2. Prove that if $2a^3 + 27c = 9ab,$ then the roots of $x^3 + ax^2 + bx + c = 0$ form an arithmetic sequence.

How should I solve this problem?

If the arithmetic sequence is of the form $(0,d,2d)$, then $$x^3+ax^2+bx+c=x(x-d)(x-2d)=x^3-3dx^2+2dx.$$ So $a=-3d$, $b=2d^2$, and $c=0$. Clearly, $$2a^3+27c=2(-3d)^3+27(0)=-54d^3$$ and $$9ab=9(-3d)(2d^2)=-54d^3.$$ So the first problem is true in this case.

Answer 1

Let $f(x):=x^3+ax^2+bx+c$. Then, $$g(x):=f\left(x-\frac{a}{3}\right)=x^3-\frac{a^2-3b}{3}x+\frac{2a^3-9ab+27c}{27}\,.$$ Note that the roots of $f(x)$ form an arithmetic progression if and only if the roots of $g(x)$ are of the form $-d$, $0$, $+d$ for some complex number $d$, which is equivalent to $$g(x)=(x+d)x(x-d)=x^3-d^2x\,.$$ That is, the roots of $f(x)$ form an arithmetic progression if and only if the constant term of $g(x)$ is $0$, which is the same as saying that $$2a^3-9ab+27c=0\,.$$

Answer 2

Let $l-t,l, l+t$ be the roots.

Then $a=l−t+l+l+t=3l \dots\dots(1)$

$b=(l−t)l+l(l+t)+(l+t)(l−t)=3l^2−t^2 \ldots \dots(2)$

$c=(l−t)l(l+t)=l(l^2−t^2)\dots \dots(3$)

Substituting for $t^2$ from (2) in (3),

$c=l^3−l(3l^2−b)=l^3−3l^3+lb=−2l^3+lb \dots \dots(4)$

Multiplying by 27 and putting $3l=a$

$27c=−2a^3+9ab$

i.e. $2a^3–9ab+27c=0$