# How do you factor and solve x^3-216=0?

Apr 2, 2018

$\implies x = \left\{6 , - 3 \pm 3 \sqrt{3} i\right\}$

#### Explanation:

${x}^{3} = 216 \implies x = 6$

$\implies \left(x - 6\right)$ is a factor

$\left(x - 6\right) \left({x}^{2} + b x + 36\right) = {x}^{3} - 216$

$\implies - 6 {x}^{2} + b {x}^{2} = 0 {x}^{2}$

$\implies b - 6 = 0 \implies b = 6$

$\implies \left(x - 6\right) \left({x}^{2} + 6 x + 36\right) = 0$

$\implies x = \left\{6 , - 3 \pm 3 \sqrt{3} i\right\}$

Apr 2, 2018

See below:

#### Explanation:

According to the fundamental theorem of Algebra, a polynomial to the degree $n$ will have $n$ roots. In our case, a cubic will have three roots.

$x = 6$ is the first solution, which is obtained by solving the equation (taking the cube root of $216$).

If$x = 6$ is a solution, then $\left(x - 6\right)$ must be a factor of this polynomial. So, if you divide this polynomial by $\left(x - 6\right)$ using algebraic division, you will get the quadratic:

${x}^{2} + 6 x + 36$

Solving this quadratic will give you the remaining two roots:

${x}^{2} + 6 x + 36 = 0$

${\left(x + 3\right)}^{2} - 9 + 36 = 0$

${\left(x + 3\right)}^{2} + 27 = 0$

$\left(x + 3\right) = \sqrt{- 27}$

(x+3)= ±3sqrt3i

x= -3±3sqrt3i

The catch here is to remember that if you take the square root of any number you get two answers(± the answer). I solved this quadratic by 'completing the square', there is also an alternative quadratic formula which can be used to solve quadratics. The non-real solutions exist in conjugate pairs (a±bi).

Therefore, the solutions are:

$x = 6 , \left(- 3 + \sqrt{3} i\right) , \left(- 3 - \sqrt{3} i\right)$