Factor q^3+27r^3

$q^{3} + 27 r^{3}$

Step 1

Rewrite $27 r^{3}$ as ${(3 r)}^{3}$ .

$q^{3} + {(3 r)}^{3}$

Step 2

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 $a = q$ and $b = 3 r$ .

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

Step 3

Simplify.

Rewrite using the commutative property of multiplication.

$(q + 3 r) (q^{2} - 1 \cdot 3 q r + {(3 r)}^{2})$

Multiply $- 1$ by $3$ .

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

Apply the product rule to $3 r$ .

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

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

$(q + 3 r) (q^{2} - 3 q r + 9 r^{2})$

Factor q^3+27r^3

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