Algebra Examples

$\frac{a^{3} + 27 b^{3}}{a^{2} - 9 b^{2}}$

Step 1

Simplify the numerator.

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Step 1.1

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

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

Step 1.2

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

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

Step 1.3

Simplify.

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Step 1.3.1

Rewrite using the commutative property of multiplication.

$\frac{(a + 3 b) (a^{2} - 1 \cdot 3 a b + {(3 b)}^{2})}{a^{2} - 9 b^{2}}$

Step 1.3.2

Multiply $- 1$ by $3$ .

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

Step 1.3.3

Apply the product rule to $3 b$ .

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

Step 1.3.4

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

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

Step 2

Simplify the denominator.

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Step 2.1

Rewrite $9 b^{2}$ as ${(3 b)}^{2}$ .

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

Step 2.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = a$ and $b = 3 b$ .

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

Step 2.3

Multiply $3$ by $- 1$ .

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

Step 3

Cancel the common factor of $a + 3 b$ .

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Step 3.1

Cancel the common factor.

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

Step 3.2

Rewrite the expression.

$\frac{a^{2} - 3 a b + 9 b^{2}}{a - 3 b}$