Precalculus Examples

$27 a^{3} - b^{6}$

Step 1

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

${(3 a)}^{3} - b^{6}$

Step 2

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

${(3 a)}^{3} - {(b^{2})}^{3}$

Step 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 $a = 3 a$ and $b = b^{2}$ .

$(3 a - b^{2}) ({(3 a)}^{2} + 3 a b^{2} + {(b^{2})}^{2})$

Step 4

Simplify.

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Step 4.1

Apply the product rule to $3 a$ .

$(3 a - b^{2}) (3^{2} a^{2} + 3 a b^{2} + {(b^{2})}^{2})$

Step 4.2

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

$(3 a - b^{2}) (9 a^{2} + 3 a b^{2} + {(b^{2})}^{2})$

Step 4.3

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

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Step 4.3.1

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

$(3 a - b^{2}) (9 a^{2} + 3 a b^{2} + b^{2 \cdot 2})$

Step 4.3.2

Multiply $2$ by $2$ .

$(3 a - b^{2}) (9 a^{2} + 3 a b^{2} + b^{4})$