Show difference between values of turning points of $ f(x) = (c-\frac{1}{c}-x)(4-3x^2) $ How would you show that the difference between the values of the turning points of
$$
f(x) = (c-\frac{1}{c}-x)(4-3x^2)
$$
is 
$$
\frac{4}{9}(c+\frac{1}{c})^3
$$
$c>0$
I have attempted to compute the derivative and set it to $0$ (which is the standard approach), but things get messy (in my opinion).

Starting with
$$f(x) = (c-\frac{1}{c}-x)(4-3x^2)$$
to make computation a bit more manageable, set $\boxed{\gamma=c-\frac{1}{c}}$, so
$$f(x)=4\gamma-4x-3\gamma x^2+3x^3$$
You had the correct idea to set $f'(x)=0$ at the turning points. So
$$f'(x)=-4-6\gamma x+9x^2=0\quad(\text{at turning points})$$
By the quadratic formula
$$x=\frac{6\gamma\pm\sqrt{36\gamma^2+144}}{18}=\frac{\gamma\pm\sqrt{\gamma^2+4}}{3}$$
At this stage, see if the quantity under the square root can be simplified:
$$\gamma^2+4=\left(c-\frac{1}{c}\right)^2+4=\left(c^2-2+\frac{1}{c^2}\right)+4=c^2+2+\frac{1}{c^2}=\left(c+\frac{1}{c}\right)^2$$
Often this sort of simplification will work as the goal of such a question is not typically to test whether you can perform messy number crunching, e.g. computing $f(1+\frac{1}{\sqrt5})$.
So the two roots are given by
$$x_{1,2}=\frac{\left(c-\frac{1}{c}\right)\pm\left(c+\frac{1}{c}\right)}{3} \implies x_1=-\frac{2}{3c},\:x_2=\frac{2c}{3}$$ 
I will omit the computation of $f(x_1)$ and $f(x_2)$.
