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.
 A: 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.
A: The function is strictly increasing so the function is one to one.
A: A solution that only uses school algebra:
Substitute $t=x-5/2$. Then the equation becomes
$$(t+3/2)^3+(t+1/2)^3+(t-1/2)^3+(t-3/2)^3=0.$$
Expanding the brackets, we get 
$$4t^3+15t=0,$$
or $t(4t^2+15)=0$,
which clearly only has one real root.
A: Hint:
Can you show the derivative is positive for all $x$?
Thus, the function is strictly increasing for all $x$.
$x=2.5$ is a root.
A: 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 $(a-b)^3+(a+b)^3=2(a^3+3ab^2)$
$$0=4y^3+6y((1.5)^2+(.5)^2)$$
A: Hint:
It is the sum of four increasing functions, and $\lim_{x\to-\infty}=-\infty$,  $\lim_{x\to+\infty}=+\infty$. The intermediate value theorem guarantees there is a root, and monotonicity ensures there can be no more than one.
A: 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, 
$$p(x) = (2x-5)\color{blue}{q(x)} \mbox{ with } \color{blue}{q(x)} > 0 \mbox{ for all } x \in \mathbb{R}$$
A: The function $x\mapsto x^3$ is increasing, since $x^3-y^3=(x-y)(x^2+xy+y^2)$ and $x^2+xy+y^2=\frac{(x+y)^2}{2}+\frac{x^2+y^2}{2}\geq 0$.
Shifting the domain of an increasing function does not change the fact that it is increasing, so also $x\mapsto (x-n)^3$ is increasing for all $n$.
Adding increasing functions results in another increasing function.
Thus, for all natural numbers $n$ the polynomial
$$
\sum_{i=1}^n (x-i)^3
$$
is increasing. Since it tends to $\infty$ as $x\to\infty$ and $-\infty$ as $x\to-\infty$, it follows that the polynomial has exactly one value at which it crosses the $x$-axis, that is, exactly one real root.
