Factor by Grouping x^3-8y^3

$x^{3} - 8 y^{3}$

Step 1

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

$x^{3} - {(2 y)}^{3}$

Step 2

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 2 y$ .

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

Step 3

Simplify.

Multiply $2$ by $- 1$ .

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

Rewrite using the commutative property of multiplication.

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

Apply the product rule to $2 y$ .

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

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

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

Factor by Grouping x^3-8y^3

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