Factor 64g^3+8

$64 g^{3} + 8$

Step 1

Factor $8$ out of $64 g^{3} + 8$ .

Factor $8$ out of $64 g^{3}$ .

$8 (8 g^{3}) + 8$

Factor $8$ out of $8$ .

$8 (8 g^{3}) + 8 (1)$

Factor $8$ out of $8 (8 g^{3}) + 8 (1)$ .

$8 (8 g^{3} + 1)$

Step 2

Rewrite $8 g^{3}$ as ${(2 g)}^{3}$ .

$8 ({(2 g)}^{3} + 1)$

Step 3

Rewrite $1$ as $1^{3}$ .

$8 ({(2 g)}^{3} + 1^{3})$

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 $a = 2 g$ and $b = 1$ .

$8 ((2 g + 1) ({(2 g)}^{2} - (2 g) \cdot 1 + 1^{2}))$

Step 5

Factor.

Simplify.

Apply the product rule to $2 g$ .

$8 ((2 g + 1) (2^{2} g^{2} - (2 g) \cdot 1 + 1^{2}))$

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

$8 ((2 g + 1) (4 g^{2} - (2 g) \cdot 1 + 1^{2}))$

Multiply $2$ by $- 1$ .

$8 ((2 g + 1) (4 g^{2} - 2 g \cdot 1 + 1^{2}))$

Multiply $- 2$ by $1$ .

$8 ((2 g + 1) (4 g^{2} - 2 g + 1^{2}))$

One to any power is one.

$8 ((2 g + 1) (4 g^{2} - 2 g + 1))$

Remove unnecessary parentheses.

$8 (2 g + 1) (4 g^{2} - 2 g + 1)$

Factor 64g^3+8

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