Let $a,b,c$ be distinct real numbers. What is the number of distinct real roots of the equation $(x-a)^3+(x-b)^3+(x-c)^3=0$?
- $1$,
- $2$,
- $3$,
- Depends on the value of $a,b,c$.
How can I solve this?
Let $a,b,c$ be distinct real numbers. What is the number of distinct real roots of the equation $(x-a)^3+(x-b)^3+(x-c)^3=0$?
How can I solve this?
Hint:
I hope this helps ;-)
From the derivative, we see that there is a sum of three squares equalling zero, $$(x-a)^2+(x-b)^2+(x-c)^2 = 0$$
which implies that for a solution to exist $a=b=c$ Since, given that they are distint, the derivative has no roots. By Rolle's theorem, we may conclude that the original function may have atmost one root.
Let $y=(x-a)^3 +(x-b)^3 +(x-c)^3$
then $\dfrac{dy}{dx} =3( (x-a)^2 +(x-b)^2 +(x-c)^2 )$
therefore, $\dfrac{dy}{dx} > 0$
Also, $f(-\infty) = -\infty$ and $f( \infty ) = \infty$
i.e., $f(x)$ is strictly increasing from $-\infty$ to $\infty$
hence the function possesses $1$ real root for any values of $a, b ,c$
the number of answers is one.