Question

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

Answer 1

`=> x = { 6 , -3 pm 3sqrt(3)i }`

Explanation:

`x^3 = 216 => x = 6`

`=> (x-6)` is a factor

`(x-6)(x^2 +bx + 36 ) = x^3 - 216`

`=> -6x^2 +bx^2 = 0x^2`

`=> b-6 = 0 => b = 6`

`=> (x-6)(x^2 + 6x + 36 ) = 0`

`=> x = { 6 , -3 pm 3sqrt(3)i }`

Using quadratic formula

Answer 2

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 `(x-6)` must be a factor of this polynomial. So, if you divide this polynomial by `(x-6)` using algebraic division, you will get the quadratic:

`x^2+6x+36`

Solving this quadratic will give you the remaining two roots:

`x^2+6x+36=0`

`(x+3)^2-9+36=0`

`(x+3)^2+27=0`

`(x+3)= 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, (-3+sqrt3i), (-3-sqrt3i)`