Simplify (b^3-8)/(b^2-9)*(b+3)/(b^2+2b+4)

$\frac{b^{3} - 8}{b^{2} - 9} \cdot \frac{b + 3}{b^{2} + 2 b + 4}$

Step 1

Simplify the numerator.

Rewrite $8$ as $2^{3}$ .

$\frac{b^{3} - 2^{3}}{b^{2} - 9} \cdot \frac{b + 3}{b^{2} + 2 b + 4}$

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

$\frac{(b - 2) (b^{2} + b \cdot 2 + 2^{2})}{b^{2} - 9} \cdot \frac{b + 3}{b^{2} + 2 b + 4}$

Simplify.

Move $2$ to the left of $b$ .

$\frac{(b - 2) (b^{2} + 2 \cdot b + 2^{2})}{b^{2} - 9} \cdot \frac{b + 3}{b^{2} + 2 b + 4}$

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

$\frac{(b - 2) (b^{2} + 2 b + 4)}{b^{2} - 9} \cdot \frac{b + 3}{b^{2} + 2 b + 4}$

Step 2

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

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

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

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

Step 3

Reduce the expression by cancelling the common factors.

Cancel the common factor of $b^{2} + 2 b + 4$ .

Factor $b^{2} + 2 b + 4$ out of $(b - 2) (b^{2} + 2 b + 4)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $b + 3$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{b - 2}{b - 3}$

Simplify (b^3-8)/(b^2-9)*(b+3)/(b^2+2b+4)

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