Factor 7x^3+56

$7 x^{3} + 56$

Step 1

Factor $7$ out of $7 x^{3} + 56$ .

Factor $7$ out of $7 x^{3}$ .

$7 (x^{3}) + 56$

Factor $7$ out of $56$ .

$7 x^{3} + 7 \cdot 8$

Factor $7$ out of $7 x^{3} + 7 \cdot 8$ .

$7 (x^{3} + 8)$

Step 2

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

$7 (x^{3} + 2^{3})$

Step 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 $b = 2$ .

$7 ((x + 2) (x^{2} - x \cdot 2 + 2^{2}))$

Step 4

Factor.

Simplify.

Multiply $2$ by $- 1$ .

$7 ((x + 2) (x^{2} - 2 x + 2^{2}))$

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

$7 ((x + 2) (x^{2} - 2 x + 4))$

Remove unnecessary parentheses.

$7 (x + 2) (x^{2} - 2 x + 4)$

Factor 7x^3+56

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