How do you factor $x^3 - 16$ ?

Question

8477 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-07T22:21:01

$x^3-16$

$=(x-2 root(3)(2))(x^2+2 root(3)(2) x + 4 root(3)(4))$

$=(x-2 root(3)(2))(x-2 root(3)(2) omega)(x-2 root(3)(2) omega^2)$

Use the difference of cubes identity:

$a^3-b^3 = (a-b)(a^2+ab+b^2)$

with $a = x$ and $b = 2root(3)(2)$

So:

$x^3-16 =x^3-(2 root(3)(2))^3$

$=(x-2 root(3)(2))(x^2+2 root(3)(2) x + 4 root(3)(4))$

This is as far as we can go with Real coefficients, but if you allow Complex coefficients then this can be further factored as:

$=(x-2 root(3)(2))(x-2 root(3)(2) omega)(x-2 root(3)(2) omega^2)$

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

How do you factor x^3 - 16x^3 - 16?