Factor q^3r^3-8 - Math solver

$q^{3} r^{3} - 8$

Rewrite $q^{3} r^{3}$ as ${(q r)}^{3}$ .

${(q r)}^{3} - 8$

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

${(q r)}^{3} - 2^{3}$

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 $a = q r$ and $b = 2$ .

$(q r - 2) ({(q r)}^{2} + q r \cdot 2 + 2^{2})$

Simplify.

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Apply the product rule to $q r$ .

$(q r - 2) (q^{2} r^{2} + q r \cdot 2 + 2^{2})$

Move $2$ to the left of $q r$ .

$(q r - 2) (q^{2} r^{2} + 2 (q r) + 2^{2})$

Remove parentheses.

$(q r - 2) (q^{2} r^{2} + 2 q r + 2^{2})$

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

$(q r - 2) (q^{2} r^{2} + 2 q r + 4)$

Factor q^3r^3-8

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