Factor over the Complex Numbers x^3-8

$x^{3} - 8$

Rewrite $8$ as $2^{3}$ .

$x^{3} - 2^{3}$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - i^{3} = (a - i) (a^{2} + a i + i^{2})$ where $a = x$ and $i = 2$ .

$(x - 2) (x^{2} + x \cdot 2 + 2^{2})$

Simplify.

Move $2$ to the left of $x$ .

$(x - 2) (x^{2} + 2 x + 2^{2})$

Raise $2$ to the power of $2$ .

$(x - 2) (x^{2} + 2 x + 4)$

Factor over the Complex Numbers x^3-8

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