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.

Here is a rather primitive solution (although judging by the contents of your question you may already know this):
By symmetry, we can see that $x=2.5$ is a real root.  We want to show that it is the only real root.  Expanding the binomial cubes, we get $$4x^3-30x^2+90x-100$$
which factors into $$2(2x-5)(x^2-5x+10).$$
As the discriminant of the quadratic factor is negative, this tells us that $x=2.5$ is the only real root.
