<s>[INST] 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$
 [/INST] 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.
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