Factor 108b^3+256

$108 b^{3} + 256$

Step 1

Factor $4$ out of $108 b^{3} + 256$ .

Factor $4$ out of $108 b^{3}$ .

$4 (27 b^{3}) + 256$

Factor $4$ out of $256$ .

$4 (27 b^{3}) + 4 (64)$

Factor $4$ out of $4 (27 b^{3}) + 4 (64)$ .

$4 (27 b^{3} + 64)$

Step 2

Rewrite $27 b^{3}$ as ${(3 b)}^{3}$ .

$4 ({(3 b)}^{3} + 64)$

Step 3

Rewrite $64$ as $4^{3}$ .

$4 ({(3 b)}^{3} + 4^{3})$

Step 4

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 $a = 3 b$ and $b = 4$ .

$4 ((3 b + 4) ({(3 b)}^{2} - (3 b) \cdot 4 + 4^{2}))$

Step 5

Factor.

Simplify.

Apply the product rule to $3 b$ .

$4 ((3 b + 4) (3^{2} b^{2} - (3 b) \cdot 4 + 4^{2}))$

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

$4 ((3 b + 4) (9 b^{2} - (3 b) \cdot 4 + 4^{2}))$

Multiply $3$ by $- 1$ .

$4 ((3 b + 4) (9 b^{2} - 3 b \cdot 4 + 4^{2}))$

Multiply $4$ by $- 3$ .

$4 ((3 b + 4) (9 b^{2} - 12 b + 4^{2}))$

Raise $4$ to the power of $2$ .

$4 ((3 b + 4) (9 b^{2} - 12 b + 16))$

Remove unnecessary parentheses.

$4 (3 b + 4) (9 b^{2} - 12 b + 16)$

Factor 108b^3+256

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