<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] 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.
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