Factor w^3+27

$w^{3} + 27$

Step 1

Rewrite $27$ as $3^{3}$ .

$w^{3} + 3^{3}$

Step 2

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 $a = w$ and $b = 3$ .

$(w + 3) (w^{2} - w \cdot 3 + 3^{2})$

Step 3

Simplify.

Multiply $3$ by $- 1$ .

$(w + 3) (w^{2} - 3 w + 3^{2})$

Raise $3$ to the power of $2$ .

$(w + 3) (w^{2} - 3 w + 9)$

Factor w^3+27

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