Question

How do you find all the real and complex roots of `x^3 + 8 = 0`?

Answer 1

`x=-2,1+-sqrt3i`

Explanation:

Split this apart using the sum of cubes identity.

`(x+2)(x^2-2x+4)=0`

Now, you know that either

`x+2=0color(white)(xxxx)"or"color(white)(xxxx)x^2-2x+4=0`

Solving the first of these gives that `x=-2`.

The other two can be found through applying the quadratic formula on the quadratic:

`x=(2+-sqrt(4-16))/2=(2+-2sqrt3i)/2=1+-sqrt3i`