Factor 192x^3-81y^3

$192 x^{3} - 81 y^{3}$

Step 1

Factor $3$ out of $192 x^{3} - 81 y^{3}$ .

Factor $3$ out of $192 x^{3}$ .

$3 (64 x^{3}) - 81 y^{3}$

Factor $3$ out of $- 81 y^{3}$ .

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

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

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

Step 2

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

$3 ({(4 x)}^{3} - 27 y^{3})$

Step 3

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

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

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

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

Step 5

Factor.

Simplify.

Multiply $3$ by $- 1$ .

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

Apply the product rule to $4 x$ .

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

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

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

Rewrite using the commutative property of multiplication.

$3 ((4 x - 3 y) (16 x^{2} + 4 \cdot 3 x y + {(3 y)}^{2}))$

Multiply $4$ by $3$ .

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

Apply the product rule to $3 y$ .

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

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

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

Remove unnecessary parentheses.

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

Factor 192x^3-81y^3

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