Solve 3/3+b-12-b/b^2-b-6-2/2-6

$$\frac{ 3 }{ 3+b } - \frac{ 12-b }{ { b }^{ 2 } -b-6 } - \frac{ 2 }{ 2-6 }$$

Evaluate

$$\frac{b^{3}+10b^{2}-33b-126}{2\left(b+2\right)\left(b^{2}-9\right)}$$

Short Solution Steps

Subtract 6 from 2 to get -4.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}-\frac{2}{-4}$$

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}-\left(-\frac{1}{2}\right)$$

The opposite of -\frac{1}{2} is \frac{1}{2}.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}+\frac{1}{2}$$

Factor b^{2}-b-6.

$$\frac{3}{3+b}-\frac{12-b}{\left(b-3\right)\left(b+2\right)}+\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3+b and \left(b-3\right)\left(b+2\right) is \left(b-3\right)\left(b+2\right)\left(b+3\right). Multiply \frac{3}{3+b} times \frac{\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)}. Multiply \frac{12-b}{\left(b-3\right)\left(b+2\right)} times \frac{b+3}{b+3}.

$$\frac{3\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}-\frac{\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Since \frac{3\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} and \frac{\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3\left(b-3\right)\left(b+2\right)-\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Do the multiplications in 3\left(b-3\right)\left(b+2\right)-\left(12-b\right)\left(b+3\right).

$$\frac{3b^{2}+6b-9b-18-12b-36+b^{2}+3b}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Combine like terms in 3b^{2}+6b-9b-18-12b-36+b^{2}+3b.

$$\frac{4b^{2}-12b-54}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(b-3\right)\left(b+2\right)\left(b+3\right) and 2 is 2\left(b-3\right)\left(b+2\right)\left(b+3\right). Multiply \frac{4b^{2}-12b-54}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} times \frac{2}{2}. Multiply \frac{1}{2} times \frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}.

$$\frac{2\left(4b^{2}-12b-54\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Since \frac{2\left(4b^{2}-12b-54\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)} and \frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{2\left(4b^{2}-12b-54\right)+\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Do the multiplications in 2\left(4b^{2}-12b-54\right)+\left(b-3\right)\left(b+2\right)\left(b+3\right).

$$\frac{8b^{2}-24b-108+b^{3}+5b^{2}+6b-3b^{2}-15b-18}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Combine like terms in 8b^{2}-24b-108+b^{3}+5b^{2}+6b-3b^{2}-15b-18.

$$\frac{10b^{2}-33b-126+b^{3}}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Expand 2\left(b-3\right)\left(b+2\right)\left(b+3\right).

$$\frac{10b^{2}-33b-126+b^{3}}{2b^{3}+4b^{2}-18b-36}$$

Expand

$$\frac{b^{3}+10b^{2}-33b-126}{2\left(b+3\right)\left(b^{2}-b-6\right)}$$

Short Solution Steps

Subtract 6 from 2 to get -4.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}-\frac{2}{-4}$$

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}-\left(-\frac{1}{2}\right)$$

The opposite of -\frac{1}{2} is \frac{1}{2}.

$$\frac{3}{3+b}-\frac{12-b}{b^{2}-b-6}+\frac{1}{2}$$

Factor b^{2}-b-6.

$$\frac{3}{3+b}-\frac{12-b}{\left(b-3\right)\left(b+2\right)}+\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3+b and \left(b-3\right)\left(b+2\right) is \left(b-3\right)\left(b+2\right)\left(b+3\right). Multiply \frac{3}{3+b} times \frac{\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)}. Multiply \frac{12-b}{\left(b-3\right)\left(b+2\right)} times \frac{b+3}{b+3}.

$$\frac{3\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}-\frac{\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Since \frac{3\left(b-3\right)\left(b+2\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} and \frac{\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3\left(b-3\right)\left(b+2\right)-\left(12-b\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Do the multiplications in 3\left(b-3\right)\left(b+2\right)-\left(12-b\right)\left(b+3\right).

$$\frac{3b^{2}+6b-9b-18-12b-36+b^{2}+3b}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

Combine like terms in 3b^{2}+6b-9b-18-12b-36+b^{2}+3b.

$$\frac{4b^{2}-12b-54}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{1}{2}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(b-3\right)\left(b+2\right)\left(b+3\right) and 2 is 2\left(b-3\right)\left(b+2\right)\left(b+3\right). Multiply \frac{4b^{2}-12b-54}{\left(b-3\right)\left(b+2\right)\left(b+3\right)} times \frac{2}{2}. Multiply \frac{1}{2} times \frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{\left(b-3\right)\left(b+2\right)\left(b+3\right)}.

$$\frac{2\left(4b^{2}-12b-54\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}+\frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Since \frac{2\left(4b^{2}-12b-54\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)} and \frac{\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{2\left(4b^{2}-12b-54\right)+\left(b-3\right)\left(b+2\right)\left(b+3\right)}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Do the multiplications in 2\left(4b^{2}-12b-54\right)+\left(b-3\right)\left(b+2\right)\left(b+3\right).

$$\frac{8b^{2}-24b-108+b^{3}+5b^{2}+6b-3b^{2}-15b-18}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Combine like terms in 8b^{2}-24b-108+b^{3}+5b^{2}+6b-3b^{2}-15b-18.

$$\frac{10b^{2}-33b-126+b^{3}}{2\left(b-3\right)\left(b+2\right)\left(b+3\right)}$$

Expand 2\left(b-3\right)\left(b+2\right)\left(b+3\right).

$$\frac{10b^{2}-33b-126+b^{3}}{2b^{3}+4b^{2}-18b-36}$$