Algebra Examples

$8 a^{3} b^{3} + 27$

Step 1

Rewrite $8 a^{3} b^{3}$ as ${(2 a b)}^{3}$ .

${(2 a b)}^{3} + 27$

Step 2

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

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

Step 3

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 $a = 2 a b$ and $b = 3$ .

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

Step 4

Simplify.

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Step 4.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 4.1.1

Apply the product rule to $2 a b$ .

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

Step 4.1.2

Apply the product rule to $2 a$ .

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

Step 4.2

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

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

Step 4.3

Multiply $2$ by $- 1$ .

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

Step 4.4

Remove parentheses.

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

Step 4.5

Multiply $3$ by $- 2$ .

$(2 a b + 3) (4 a^{2} b^{2} - 6 a b + 3^{2})$

Step 4.6

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

$(2 a b + 3) (4 a^{2} b^{2} - 6 a b + 9)$