# How do you solve x^3-6x^2+6x=0?

Dec 17, 2015

I found:
${x}_{1} = 0$
${x}_{2} = 3 + \sqrt{3}$
${x}_{3} = 3 - \sqrt{3}$

#### Explanation:

We can try collecting $x$:
$x \left({x}^{2} - 6 x + 6\right) = 0$
you get the first solution from the $x \left(\ldots\right) = 0$ bit which gives you: ${x}_{1} = 0$
the other two solutions are found solving the ${x}^{2} - 6 x + 6 = 0$ bit using the Quadratic Formula:
${x}_{2 , 3} = \frac{6 \pm \sqrt{36 - 24}}{2} = \frac{6 \pm \sqrt{12}}{2} = \frac{6 \pm 2 \sqrt{3}}{2} = 3 \pm \sqrt{3}$
giving:
${x}_{2} = 3 + \sqrt{3}$
${x}_{3} = 3 - \sqrt{3}$