Pre-Algebra Examples

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

Step 1

Simplify the numerator.

Tap for more steps...

Step 1.1

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

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

Step 1.2

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 $b = 2$ .

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

Step 1.3

Simplify.

Tap for more steps...

Step 1.3.1

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

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

Step 1.3.2

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

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

Step 2

Simplify terms.

Tap for more steps...

Step 2.1

Add $8$ and $3$ .

$\frac{(b - 2) (b^{2} + 2 b + 4)}{11} \cdot \frac{b + 2}{4 + 2 b + b^{2}}$

Step 2.2

Multiply $\frac{(b - 2) (b^{2} + 2 b + 4)}{11}$ by $\frac{b + 2}{4 + 2 b + b^{2}}$ .

$\frac{(b - 2) (b^{2} + 2 b + 4) (b + 2)}{11 (4 + 2 b + b^{2})}$

Step 2.3

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

Tap for more steps...

Step 2.3.1

Reorder terms.

$\frac{(b - 2) (b^{2} + 2 b + 4) (b + 2)}{11 (b^{2} + 2 b + 4)}$

Step 2.3.2

Cancel the common factor.

$\frac{(b - 2) (b^{2} + 2 b + 4) (b + 2)}{11 (b^{2} + 2 b + 4)}$

Step 2.3.3

Rewrite the expression.

$\frac{(b - 2) (b + 2)}{11}$