<s>[INST] 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.
 [/INST] 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, 
$$p(x) = (2x-5)\color{blue}{q(x)} \mbox{ with } \color{blue}{q(x)} > 0 \mbox{ for all } x \in \mathbb{R}$$
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