Finding the number of distinct real roots of $(x-a)^3+(x-b)^3+(x-c)^3=0$ for $a$, $b$, $c$ distinct real numbers

Question

This question was asked to me by a junior and I was unable to solve it. So, I am asking for help here.

Let $a$, $b$, $c$ be distinct real numbers. Then the number of distinct real roots of the equation $(x-a)^3+(x-b)^3+(x-c)^3=0$ is

(i) $1\quad$ (ii) $2\quad$ (iii) $3$

Answer

(i) $1$

Well , in the question field is not given but even if it was given i don't know how to approach this particular question.

I have studied solution of cubic equations 4 years back but I am asking if it's possible to solve it without use results from solutions of cubic equation, as they are very rarely used and hence I don't remember them.

Thanks!!

Answer 1

$$f={(x-a)^3}+{(x-b)^3}+{(x-c)^3}$$

$$f'={3(x-a)^2}+{3(x-b)^2}+{3(x-c)^2}\ge0$$

This shows that the graph is non decreasing. So $f$ crosses x-axis only once

Note that the equality occurs when $x=a=b=c=t$