Factor x^3+8

$x^{3} + 8$

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

$x^{3} + 2^{3}$

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

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

Simplify.

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Multiply $2$ by $- 1$ .

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

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

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

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