Algebra Examples

$r^{6} s^{6} + 27 s^{3}$

Step 1

Factor $s^{3}$ out of $r^{6} s^{6} + 27 s^{3}$ .

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Step 1.1

Factor $s^{3}$ out of $r^{6} s^{6}$ .

$s^{3} (r^{6} s^{3}) + 27 s^{3}$

Step 1.2

Factor $s^{3}$ out of $27 s^{3}$ .

$s^{3} (r^{6} s^{3}) + s^{3} \cdot 27$

Step 1.3

Factor $s^{3}$ out of $s^{3} (r^{6} s^{3}) + s^{3} \cdot 27$ .

$s^{3} (r^{6} s^{3} + 27)$

Step 2

Rewrite $r^{6} s^{3}$ as ${(r^{2} s)}^{3}$ .

$s^{3} ({(r^{2} s)}^{3} + 27)$

Step 3

Rewrite $27$ as $3^{3}$ .

$s^{3} ({(r^{2} s)}^{3} + 3^{3})$

Step 4

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 $a = r^{2} s$ and $b = 3$ .

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

Step 5

Factor.

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Step 5.1

Simplify.

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Step 5.1.1

Apply the product rule to $r^{2} s$ .

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

Step 5.1.2

Multiply the exponents in ${(r^{2})}^{2}$ .

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Step 5.1.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 5.1.2.2

Multiply $2$ by $2$ .

$s^{3} ((r^{2} s + 3) (r^{4} s^{2} - (r^{2} s) \cdot 3 + 3^{2}))$

Step 5.1.3

Remove parentheses.

$s^{3} ((r^{2} s + 3) (r^{4} s^{2} - r^{2} s \cdot 3 + 3^{2}))$

Step 5.1.4

Multiply $3$ by $- 1$ .

$s^{3} ((r^{2} s + 3) (r^{4} s^{2} - 3 r^{2} s + 3^{2}))$

Step 5.1.5

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

$s^{3} ((r^{2} s + 3) (r^{4} s^{2} - 3 r^{2} s + 9))$

Step 5.2

Remove unnecessary parentheses.

$s^{3} (r^{2} s + 3) (r^{4} s^{2} - 3 r^{2} s + 9)$