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 an elementary way that uses only 


*

*$(1)$: $a^3+b^3 = (a+b)(a^2-ab+b^2)$ and

*$(2)$: $a^2+b^2 >ab$ for $|a|+|b| > 0$
Note, that $|x-1|+|x-4|\geq |x-1 +(4-x)| =3 > 0$ and $|x-2|+|x-3|\geq |x-2 +(3-x)| =1 > 0$
Let's call $p(x) = (x-1)^3+ (x-2)^3 + (x-3)^3 + (x-4)^3$.
Now, using $(1)$ write
$$(x-1)^3 + (x-4)^3 = (x-1 + x-4)\left((x-1)^2 + (x-4)^2 - (x-1)(x-4)\right)$$
$$= (2x-5)\left((x-1)^2 + (x-4)^2 - (x-1)(x-4)\right)$$
$$(x-2)^3 + (x-3)^3 = (x-2 + x-3)\left((x-2)^2 + (x-3)^2 - (x-2)(x-3)\right)$$ $$= (2x-5)\left((x-2)^2 + (x-3)^2 - (x-2)(x-3)\right)$$
Hence,
$$p(x) =  $$
$$(2x-5)\color{blue}{\left(\underbrace{(x-1)^2 + (x-4)^2 - (x-1)(x-4)}_{\stackrel{(2)}{>}0} + \underbrace{(x-2)^2 + (x-3)^2 - (x-2)(x-3)}_{\stackrel{(2)}{>}0}\right)}
$$
So, 
$$p(x) = (2x-5)\color{blue}{q(x)} \mbox{ with } \color{blue}{q(x)} > 0 \mbox{ for all } x \in \mathbb{R}$$
