If the zeros of a cubic form an arithmetic sequence, its point of inflection lies on the $x$-axis

Question

If the zeros of the cubic $$y=x^3+ax^2+bx+c $$ form an arithmetic sequence, then show that the point of inflexion lies on the $x$-axis.

Answer 1

A $3$rd-degree polynomial cannot have more than three zeroes, so the zeroes must be $r$, $r+d$, and $r-d$. Hence \begin{align} & x^3 + ax^2+bx+c = (x-r)\Big(x-(r+d)\Big)\Big(x-(r-d)\Big) \\[10pt] = {} & x^3 -3rx^2 + (3r^2-d^2) x + rd^2-r^3. \end{align} Therefore $$ y'' = 6(x-r) $$ and you have an inflection point at $r$ which is also a zero of the original function.

Answer 2

Your zeroes are $x_0,x_0+d,x_0+2d$, as they form an arithmetic sequence. Factoring your cubic gives you

$$(x-x_0)(x-x_0-d)(x-x_0-2d)=(x-x_0)^3 - 3d(x-x_0)^2 + 2d^2(x-x_0).$$

The second derivative of this cubic is

$$6(x-x_0) - 6d,$$

and if you equate this to $0$, you get $x = x_0+d$, which you know from the beginning was a $0$ of the cubic.