Prove that the equation has only one real root. Prove that $(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=0$ has only one real root.
It's easy to show that the equation has a real root using Rolle's theorem. But how to show that the real root is unique? By Descartes' rule of sign, it can be shown that it has 3 or 1 real root. 
But it doesn't guarantee that the real root is unique. If we calculate the root then it can be shown that it has only one real root.

Let $y=\dfrac{x-1+x-2+x-3+x-4}4$
$x=y+2.5$
$$(x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=(y+1.5)^3+(y+.5)^3+(y-.5)^3+(y-1.5)^3$$
Now use $(a-b)^3+(a+b)^3=2(a^3+3ab^2)$
$$0=4y^3+6y((1.5)^2+(.5)^2)$$
