Factor 54x^3-128y^3

$54 x^{3} - 128 y^{3}$

Step 1

Factor $2$ out of $54 x^{3} - 128 y^{3}$ .

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

$2 (27 x^{3}) - 128 y^{3}$

Factor $2$ out of $- 128 y^{3}$ .

$2 (27 x^{3}) + 2 (- 64 y^{3})$

Factor $2$ out of $2 (27 x^{3}) + 2 (- 64 y^{3})$ .

$2 (27 x^{3} - 64 y^{3})$

Step 2

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

$2 ({(3 x)}^{3} - 64 y^{3})$

Step 3

Rewrite $64 y^{3}$ as ${(4 y)}^{3}$ .

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

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 $a = 3 x$ and $b = 4 y$ .

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

Step 5

Factor.

Simplify.

Multiply $4$ by $- 1$ .

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

Apply the product rule to $3 x$ .

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

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

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

Rewrite using the commutative property of multiplication.

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

Multiply $3$ by $4$ .

$2 ((3 x - 4 y) (9 x^{2} + 12 x y + {(4 y)}^{2}))$

Apply the product rule to $4 y$ .

$2 ((3 x - 4 y) (9 x^{2} + 12 x y + 4^{2} y^{2}))$

Raise $4$ to the power of $2$ .

$2 ((3 x - 4 y) (9 x^{2} + 12 x y + 16 y^{2}))$

Remove unnecessary parentheses.

$2 (3 x - 4 y) (9 x^{2} + 12 x y + 16 y^{2})$

Factor 54x^3-128y^3

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