Proving that extrema of cubic with 3 distinct roots always happen to fall between the roots By Rolle's Theorem, it is possible to prove that between points $a$ and $b$ there is a point $c$ at which the value of $f'(c)=0$. 
Now, consider a cubic polynomial function with 3 distinct real roots,
$f(x)=A(x-a)(x-b)(x-c)$
It is now necessary to prove that $x$-coordinates of extrema of $f(x)$ fall between $a, b$ and $c$. Or otherwise stated,
$a<x_1<b<x_2<c$, where $x_1$ and $x_2$ are $x$-coordinates of extrema.
The problem is, however, in the fact that this should be proven without the usage of Rolle's Theorem. I can prove this fact for quadratic, I get $x=(a+b)/2$, which suggests that extremum is halfway between the roots. With cubic, however, it is much more difficult. Any ideas?

I'm taking capital $A=1$ and then $a < b < c$ as the roots. we get the cubic 
$$ x^3 - \sigma_1 x^2 + \sigma_2 x - \sigma_3 \; , $$
where
$$ \sigma_1 = a + b + c, $$
$$ \sigma_2 = bc + ca + ab, $$
$$ \sigma_3 = abc. $$
The first derivative is $3 x^2 - 2 \sigma_1 x + \sigma_2,$ with roots
$$ \frac{\sigma_1 \pm \sqrt{\sigma_1^2 - 3 \sigma_2}}{3} \; .$$
Worth emphasizing that, with $a,b,c$ distinct, we get 
$$ \sigma_1^2 - 3 \sigma_2 = a^2 + b^2 + c^2 - bc - ca - ab = \frac{1}{2} \left( (b-c)^2 + (c-a)^2 + (a-b)^2 \right)  $$
being strictly positive, so the two roots of $3 x^2 - 2 \sigma_1 x + \sigma_2$ are real and distinct.
The claim that $c > \frac{\sigma_1 + \sqrt{\sigma_1^2 - 3 \sigma_2}}{3}$ comes to combining the observation that $c > \frac{\sigma_1}{3}$ along with 
$$ \left( c - \frac{\sigma_1}{3} \right)^2 -\left(  \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-a)(c-b) > 0 $$
The claim that $a < \frac{\sigma_1 - \sqrt{\sigma_1^2 - 3 \sigma_2}}{3}$ comes to combining the observation that $a < \frac{\sigma_1}{3}$ along with 
$$ \left( a - \frac{\sigma_1}{3} \right)^2 -\left(  \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-a)(b-a) > 0 $$
The final claim is that $b$ lies between the critical points, in that the distance between $b$ and $\sigma_1/3$ is smaller than $\sqrt{\sigma_1^2 - 3 \sigma_2}/3.$ Indeed,
 $$ \left( b - \frac{\sigma_1}{3} \right)^2 -\left(  \frac{\sigma_1^2 - 3 \sigma_2}{9} \right) = \frac{1}{3} (c-b)(a-b) < 0 $$
