Factor by Grouping 64x^3+8

$64 x^{3} + 8$

Step 1

Factor out the GCF of $8$ from $64 x^{3} + 8$ .

Factor out the GCF of $8$ from each term in the polynomial.

Factor out the GCF of $8$ from the expression $64 x^{3}$ .

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

Factor out the GCF of $8$ from the expression $8$ .

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

Since all the terms share a common factor of $8$ , it can be factored out of each term.

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

Step 2

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

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

Step 3

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

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

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

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

Step 5

Simplify.

Apply the product rule to $2 x$ .

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

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

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

Multiply $2$ by $- 1$ .

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

Multiply $- 2$ by $1$ .

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

One to any power is one.

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

Factor by Grouping 64x^3+8

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