Question

How do you factor completely `a^3 - 8`?

Answer 1

`a^3-8 = (a-2)(a^2+2a+4)`

`=(a-2)(a+1-sqrt(3)i)(a+1+sqrt(3)i)`

Explanation:

Use the difference of cubes identity, which can be written:

`A^3-B^3=(A-B)(A^2+AB+B^2)`

Note that both `a^3` and `8=2^3` are both perfect cubes.

So we find:

`a^3-8`

`=a^3-2^3`

`=(a-2)(a^2+(a)(2)+2^2)`

`=(a-2)(a^2+2a+4)`

The remaining quadratic factor has negative discriminant, so no Real zeros and no linear factors with Real coefficients. It is possible to factor it using Complex coefficients:

`a^2+2a+4`

`=a^2+2a+1+3`

`=(a+1)^2+3`

`=(a+1)^2-(sqrt(3)i)^2`

`= (a+1-sqrt(3)i)(a+1+sqrt(3)i)`

So:

`a^8-8 = (a-2)(a+1-sqrt(3)i)(a+1+sqrt(3)i)`

Another way to express the full factoring is:

`a^3-8 = (a-2)(a-2omega)(a-2omega^2)`

where `omega = -1/2+sqrt(3)/2 i` is the primitive Complex cube root of `1`.