The problem is to show that the equation $\displaystyle (x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=0$ has only 1 real root.

After simplification, the equation comes out to be of the form $\displaystyle ax^3-bx^2+cx-d=0$. Applying Descarte's rule of sign we get that the number of +ve real root is 1 or 3, and there is no -ve real root. How can we say then that the number of real root is exactly 1?