Factor 16x^3-54y^3

$16 x^{3} - 54 y^{3}$

Step 1

Factor $2$ out of $16 x^{3} - 54 y^{3}$ .

Factor $2$ out of $16 x^{3}$ .

$2 (8 x^{3}) - 54 y^{3}$

Factor $2$ out of $- 54 y^{3}$ .

$2 (8 x^{3}) + 2 (- 27 y^{3})$

Factor $2$ out of $2 (8 x^{3}) + 2 (- 27 y^{3})$ .

$2 (8 x^{3} - 27 y^{3})$

Step 2

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

$2 ({(2 x)}^{3} - 27 y^{3})$

Step 3

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

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

Step 4

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 $a = 2 x$ and $b = 3 y$ .

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

Step 5

Factor.

Simplify.

Multiply $3$ by $- 1$ .

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

Apply the product rule to $2 x$ .

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

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

$2 ((2 x - 3 y) (4 x^{2} + 2 x (3 y) + {(3 y)}^{2}))$

Rewrite using the commutative property of multiplication.

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

Multiply $2$ by $3$ .

$2 ((2 x - 3 y) (4 x^{2} + 6 x y + {(3 y)}^{2}))$

Apply the product rule to $3 y$ .

$2 ((2 x - 3 y) (4 x^{2} + 6 x y + 3^{2} y^{2}))$

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

$2 ((2 x - 3 y) (4 x^{2} + 6 x y + 9 y^{2}))$

Remove unnecessary parentheses.

$2 (2 x - 3 y) (4 x^{2} + 6 x y + 9 y^{2})$

Factor 16x^3-54y^3

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