Factor 16x^3-250y^3 - MathMaster

$16 x^{3} - 250 y^{3}$

Factor $2$ out of $16 x^{3} - 250 y^{3}$ .

Tap for more steps…

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

$2 (8 x^{3}) - 250 y^{3}$

Factor $2$ out of $- 250 y^{3}$ .

$2 (8 x^{3}) + 2 (- 125 y^{3})$

Factor $2$ out of $2 (8 x^{3}) + 2 (- 125 y^{3})$ .

$2 (8 x^{3} - 125 y^{3})$

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

$2 ({(2 x)}^{3} - 125 y^{3})$

Rewrite $125 y^{3}$ as ${(5 y)}^{3}$ .

$2 ({(2 x)}^{3} - {(5 y)}^{3})$

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 $b = 5 y$ .

$2 ((2 x - (5 y)) ({(2 x)}^{2} + 2 x (5 y) + {(5 y)}^{2}))$

Factor.

Tap for more steps…

Simplify.

Tap for more steps…

Multiply $5$ by $- 1$ .

$2 ((2 x - 5 y) ({(2 x)}^{2} + 2 x (5 y) + {(5 y)}^{2}))$

Apply the product rule to $2 x$ .

$2 ((2 x - 5 y) (2^{2} x^{2} + 2 x (5 y) + {(5 y)}^{2}))$

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

$2 ((2 x - 5 y) (4 x^{2} + 2 x (5 y) + {(5 y)}^{2}))$

Rewrite using the commutative property of multiplication.

$2 ((2 x - 5 y) (4 x^{2} + 2 \cdot 5 x y + {(5 y)}^{2}))$

Multiply $2$ by $5$ .

$2 ((2 x - 5 y) (4 x^{2} + 10 x y + {(5 y)}^{2}))$

Apply the product rule to $5 y$ .

$2 ((2 x - 5 y) (4 x^{2} + 10 x y + 5^{2} y^{2}))$

Raise $5$ to the power of $2$ .

$2 ((2 x - 5 y) (4 x^{2} + 10 x y + 25 y^{2}))$

Remove unnecessary parentheses.

$2 (2 x - 5 y) (4 x^{2} + 10 x y + 25 y^{2})$

Factor 16x^3-250y^3

Download our
App from the store

Create a High Performed UI/UX Design from a Silicon Valley.

Download our App from the store

Create a High Performed UI/UX Design from a Silicon Valley.

Download our
App from the store