Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ 
Problem:Find the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ where $a,b,c$ are distinct real numbers
Solution:$(x-a)^3+(x-b)^3+(x-c)^3=0$
$3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-a^3-b^3-c^3=0$
By Descartes rule of sign,number of positive real roots $=3$

But are they distinct $?$
Answer :- number of distinct real roots $ =1$
 A: If $$f(x)=(x-a)^3+(x-b)^3+(x-c)^3$$ then $$f'(x)=3(x-a)^2+3(x-b)^2+3(x-c)^2$$ Since $a,b,c$ are distinct real numbers $f'(x) > 0$ for all $x\in\mathbb{R}$ and therefore $f$ is strictly increasing and therefore it has only one real root.
EDIT: The last statement is true since $f$ is a polynomial function of degree $3$ ($a_0>0$) so $\lim_{x\to-\infty} f(x) = -\infty$, $\lim_{x\to\infty} f(x) = \infty$ and $f$ is continuous.
A: Set $m=(a+b+c)/3$, $A=a-m$, $B=b-m$, $C=c-m$ and $x=y+m$. Then your equation becomes
$$
(y-A)^3 + (y-B)^3 + (y-C)^3 = 0
$$
and, since $A+B+C=0$, your expansion applies to give
$$
y^3+(A^2+B^2+C^2)-\frac{A^3+B^3+C^3}{3}=0
$$
which is a suppressed cubic, whose discriminant is
$$
\frac{1}{4}\biggl(-\frac{A^3+B^3+C^3}{3}\biggr)^2+\frac{1}{27}(A^2+B^2+C^2)^3>0
$$
so the equation has only one real root.
A: Let's substitute the variable: $x=y+\frac{(a+b+c)}{3}$
The equation will now look like 
$$3y^3+(2a^2-2ab-2ac+2b^2-2bc+2c^2)y+\frac{(a+b-2c)(a-3b+c)(b-2a+c)}{9}=0$$
Now we apply Cardano's method.
$$Q=\left(\frac{2a^2-2ab-2ac+2b^2-2bc+2c^2}{3}\right)^3+\left(\frac{\frac{(a+b-2c)(a-3b+c)(b-2a+c)}{9}}{2}\right)^2=\frac{8}{27}(a^2-ab-ac+b^2-bc+c^2)^3+\frac{1}{324}(a+b-2c)^2(a-2b+c)^2(b-2a+c)^2$$
Then the answer is given depending on the sign of Q.
If $Q>0$ there is only one real root
If $Q<0$ there are three distinct real roots
If $Q=0$ there are two distinct real roots.
