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Given that $a,b,c$ be three distinct real numbers then the number of distinct real roots of the equation $p(x)=(x-a)^3+(x-b)^3+(x-c)^3=0$ is

  1. 1

  2. 2

  3. 3

  4. depends on $a,b,c$

what I did is $p'(x)=0$ which is two degree polynomial with three distinct root, so $p'(x)\equiv 0$ so $p(x)=k$ some constant, where I am wrong? Is the probelm wrong? Thank you for help.

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In fact, the derivative of $p(x)$ here has no real roots. –  Milind Jul 1 '13 at 14:26

4 Answers 4

up vote 3 down vote accepted

Hint: Look at at $p'(x)$ (is it always positive?) and consider $p(x)$ as $x\to-\infty$ and as $x\to\infty$.

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$p'(x)\ge 0$ and considering limits at both ends I get atleast one real root.so The answer is $1$ –  La Belle Noiseuse Jul 1 '13 at 14:35
@TaxiDriver: If there were two or more real roots, $p'(x)=0$ for some $x$ between the roots. That finishes it off. –  robjohn Jul 1 '13 at 14:36
if there is two distinct roots say $x_1,x_2$ then by rolles theorem $p'(s)=0$ for some $s\in(x_1,x_2)$ –  La Belle Noiseuse Jul 1 '13 at 14:37
@TaxiDriver: precisely! :-) –  robjohn Jul 1 '13 at 14:37
@TaxiDriver: It is good that you only got hints, but if an exam is ongoing, you should tag appropriately; either homework or contest-math. –  robjohn Jul 1 '13 at 14:42

Differentiating $p(x)$:

$$p'(x) = 3[(x-a)^2+(x-b)^2+(x-c)^2]$$

This has no real roots (Think about why not)... can you go from here?

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wait, is that statement is true- if sum of square of some expression are $0$ then must they will separately $0$? –  La Belle Noiseuse Jul 1 '13 at 14:30
@TaxiDriver: each term is $\ge0$. How can such numbers sum to $0$? –  robjohn Jul 1 '13 at 14:32
@robjohn I see! –  La Belle Noiseuse Jul 1 '13 at 14:33

You don't need to differentiate.

The sum has three components which are cubes.

Each component is strictly increasing with $x$ (all you need to do is to show that the cube function is increasing, and this is elementary), so the whole function is strictly increasing with $x$.

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You have two cases. If a = b = c then your polynomial becomes p(x) = (x-a)^3, which has a triple, real root at a. If they are not all equal than p'(x) is positive everywhere, as robjohn points out, so p(x) is forced to have just one real root. That means the other two roots are complex conjugates and not equal.

It is, of course, possible for a 3rd degree polynomial to have 3 real, distinct roots. We see that such a polynomial cannot be of the form you suggest, because its derivative must be negative somewhere.

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