Factor w^3-8

$w^{3} - 8$

Step 1

Rewrite $8$ as $2^{3}$ .

$w^{3} - 2^{3}$

Step 2

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 $a = w$ and $b = 2$ .

$(w - 2) (w^{2} + w \cdot 2 + 2^{2})$

Step 3

Simplify.

Move $2$ to the left of $w$ .

$(w - 2) (w^{2} + 2 w + 2^{2})$

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

$(w - 2) (w^{2} + 2 w + 4)$

Factor w^3-8

Related Questions

Support LCMGCF.com by disabling your adblocker.