Factor by Grouping a^3b^3-1

$a^{3} b^{3} - 1$

Rewrite $a^{3} b^{3}$ as ${(a b)}^{3}$ .

${(a b)}^{3} - 1$

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

${(a b)}^{3} - 1^{3}$

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 $a = a b$ and $b = 1$ .

$(a b - 1) ({(a b)}^{2} + a b \cdot 1 + 1^{2})$

Simplify.

Apply the product rule to $a b$ .

$(a b - 1) (a^{2} b^{2} + a b \cdot 1 + 1^{2})$

Multiply $a$ by $1$ .

$(a b - 1) (a^{2} b^{2} + a b + 1^{2})$

One to any power is one.

$(a b - 1) (a^{2} b^{2} + a b + 1)$

Factor by Grouping a^3b^3-1

Related Questions

Support LCMGCF.com by disabling your adblocker.