to find number of distinct root of a three degree polynomial 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

*2

*3

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