How many real roots does $(x-a)^3+(x-b)^3+(x-c)^3$ have?

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?

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

Hint:

• 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 ;-)

• yeap, that's the key, and all that is needed 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.

• Continuing the answer: "...and since the function is a real polynomial one of odd degree then it has one solution" Jul 16, 2013 at 10:15
• @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 Jul 16, 2013 at 10:17
• 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 Jul 16, 2013 at 17:03
• @SurajM.S I have made a new explanation in my comment, please refer. I was talking about the roots of the derivative. 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.

• This was the most succinct answer to this question I've seen. Up vote! Jul 16, 2013 at 18:11