How to prove the coordinate of an inflection point of the function $f(x)= ax^3+bx^2+cx+d$ 
Prove that the function $f(x)= ax^3+bx^2+cx+d$ has always an inflection point.

I already did that, but what I do not know how to do the second part:

If it's graphic has 3 intersections with x, say $(x_1,0),(x_2,0),(x_3,0)$, prove that the coordinate $x$ of that inflection point is $\frac{x_1+x_2+x_3}{3}$


If it has three roots, then the function $f(x)$ can be represented as 
$$f(x) = k(x-x_1)(x-x_2)(x-x_3) = k(x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_1x_3+x_2x_3)x-x_1x_2x_3)$$
Equating coefficients, you find $k=a$.
Taking the second derivative and setting equal to zero, you have:
$$f''(x) = 6ax-2a(x_1+x_2+x_3) = 0$$
$$x = \dfrac{x_1+x_2+x_3}{3}$$
