Answer
The solutions are $x = 0, 3 - \sqrt {3}, \text{ and } 3 + \sqrt {3}$.
Work Step by Step
Rewrite this equation so all the terms are on the left side of the equation and that the equation equals $0$.
$x^3 - 6x^2 + 6x = 0$
Factor out $x$:
$x(x^2 - 6x + 6) = 0$
Use the Zero-Product Property by equating each factor to $0$, then solve each equation.:
First factor:
$x = 0$
Second factor:
$x^2 - 6x + 6 = 0$
We cannot factor this polynomial, so we resort to using the quadratic formula, which is:
$x = \dfrac{-b \pm \sqrt {b^2 - 4ac}}{2a}$
where $a=1, b=-6,$ and $c=6$.
Let us plug in the numbers into the formula:
$x = \dfrac{-(-6) \pm \sqrt {(-6)^2 - 4(1)(6)}}{2(1)}$
$x = \dfrac{6 \pm \sqrt {36 - 24}}{2}$
$x = \dfrac{6 \pm \sqrt {12}}{2}$
The number $12$ can be expanded into the factors $4$ and $3$:
$x = \dfrac{6 \pm \sqrt {3(4)}}{2}$
$x = \dfrac{6 \pm 2\sqrt {3}}{2}$
$x = 3 \pm \sqrt {3}$
The solutions are $x = 0, 3 \pm \sqrt {3}$.
