1. ## problem

I have a problem but i don't know what method i should use to solve it.

If x^3 – 2x^2 + 3x = 6 prove x=2

Apologies for the poor format of the equation.

2. subtract 6 from both sides and factorize the left side:
$x^3-2x^2+3x-6 = x^2(x-2) + 3(x-2) = (x^2+3)(x-2)$
Assuming x should be a real number, this can equal to zero if and only if at least one of the brackets is zero and this happens if and only if x=2 (first bracket is never zero). In complex numbers there would be also other solutions, not only x=2.

3. Originally Posted by Maths Stig
I have a problem but i don't know what method i should use to solve it.

If x^3 – 2x^2 + 3x = 6 prove x=2

Apologies for the poor format of the equation.
We will solve this using the fact that $x^3 -2x^2 +3x - 6 = (x-2)(x^2+3)$

If we assume $x \in \mathbb{R}$ (where $\mathbb{R}$ are the real numbers), then the solution for equation would be either the solution for $x-2=0$ or $x^2 + 3 = 0$ . Obviously, $x=2$ is a solution, and there are no other solutions for $x^2 + 3 = 0$ in $\mathbb{R}$, thus $x=2$

If you want to solve this equation over $\mathbb{C}$ (the complex numbers), then $\sqrt{3}i$ , $-\sqrt{3}i$ are also solutions, but I don't think this is the case.