1. ## Theory of equation

The problem is to show that the equation $\displaystyle (x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=0$ has only 1 real root.

After simplification, the equation comes out to be of the form $\displaystyle ax^3-bx^2+cx-d=0$. Applying Descarte's rule of sign we get that the number of +ve real root is 1 or 3, and there is no -ve real root. How can we say then that the number of real root is exactly 1?

2. Originally Posted by Sambit
The problem is to show that the equation $\displaystyle (x-1)^3+(x-2)^3+(x-3)^3+(x-4)^3=0$ has only 1 real root.

After simplification, the equation comes out to be of the form $\displaystyle ax^3-bx^2+cx-d=0$. Applying Descarte's rule of sign we get that the number of +ve real root is 1 or 3, and there is no -ve real root. How can we say then that the number of real root is exactly 1?

f '(x) is always positive, so the graph of the function always raises, it means we have only one real root.

3. Originally Posted by ahaok
f '(x) is always positive, so the graph of the function always raises, it means we have only one real root.
That answer is of course correct, but in this case not useful.
This is a pre-university pre-algebra/algebra forum.
So if the question is correctly placed, then derivatives are not used.

4. Originally Posted by Plato
That answer is of course correct, but in this case not useful.
This is a pre-university pre-algebra/algebra forum.
So if the question is correctly placed, then derivatives are not used.
That's no problem. I know calculus. I posted the question here since it concerns algebra.