Factor 3+81x^3y^3

$3 + 81 x^{3} y^{3}$

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

Factor $3$ out of $3$ .

$3 (1) + 81 x^{3} y^{3}$

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

$3 (1) + 3 (27 x^{3} y^{3})$

Factor $3$ out of $3 (1) + 3 (27 x^{3} y^{3})$ .

$3 (1 + 27 x^{3} y^{3})$

Rewrite $1$ as $1^{3}$ .

$3 (1^{3} + 27 x^{3} y^{3})$

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

$3 (1^{3} + {(3 x y)}^{3})$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 1$ and $b = 3 x y$ .

$3 ((1 + 3 x y) (1^{2} - 1 (3 x y) + {(3 x y)}^{2}))$

Factor.

Simplify.

One to any power is one.

$3 ((1 + 3 x y) (1 - 1 (3 x y) + {(3 x y)}^{2}))$

Multiply $3$ by $- 1$ .

$3 ((1 + 3 x y) (1 - 3 (x y) + {(3 x y)}^{2}))$

Remove parentheses.

$3 ((1 + 3 x y) (1 - 3 x y + {(3 x y)}^{2}))$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $3 x y$ .

$3 ((1 + 3 x y) (1 - 3 x y + {(3 x)}^{2} y^{2}))$

Apply the product rule to $3 x$ .

$3 ((1 + 3 x y) (1 - 3 x y + 3^{2} x^{2} y^{2}))$

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

$3 ((1 + 3 x y) (1 - 3 x y + 9 x^{2} y^{2}))$

Remove unnecessary parentheses.

$3 (1 + 3 x y) (1 - 3 x y + 9 x^{2} y^{2})$

Factor 3+81x^3y^3

Related Questions

Support LCMGCF.com by disabling your adblocker.