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. $1$,
  2. $2$,
  3. $3$,
  4. Depends on the value of $a,b,c$.

How can I solve this?

  • 2
    $\begingroup$ Did you try to think of Rolle's theorem? $\endgroup$ Jul 16, 2013 at 8:36
  • $\begingroup$ @Santosh are you trying to solve NET's question paper? $\endgroup$
    – Mathronaut
    Jul 16, 2013 at 15:33
  • $\begingroup$ neeraj may be you thought $\endgroup$
    – sam
    Jul 17, 2013 at 4:04

3 Answers 3



  • Function $f(x) = x^3$ is strictly increasing in $x$.
  • Let $f$ and $g$ be strictly increasing, then $x \mapsto f(x)+g(x)$ is strictly increasing as well.

I hope this helps ;-)

  • 1
    $\begingroup$ yeap, that's the key, and all that is needed $\endgroup$
    – scibuff
    Jul 16, 2013 at 8:43

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.

  • 3
    $\begingroup$ Continuing the answer: "...and since the function is a real polynomial one of odd degree then it has one solution" $\endgroup$
    – DonAntonio
    Jul 16, 2013 at 10:15
  • $\begingroup$ @santoshkumar if you have a polynomial of degree n, then the derivative will be of degree n-1 and hence can admit at most n-1 roots, since the derivative has no roots then the polynomial should be a continuously increasing one(derivative is the slope, right?). Hence, only one root over reals $\endgroup$ Jul 16, 2013 at 10:17
  • $\begingroup$ i just dont understand by what he means by the existance of a root only if a=b=c. the correct answer is given by me please verify $\endgroup$
    – Suraj M S
    Jul 16, 2013 at 17:03
  • $\begingroup$ @SurajM.S I have made a new explanation in my comment, please refer. I was talking about the roots of the derivative. $\endgroup$ Jul 16, 2013 at 17:17

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.

  • 1
    $\begingroup$ This was the most succinct answer to this question I've seen. Up vote! $\endgroup$
    – Andrew
    Jul 16, 2013 at 18:11

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