Solve b^2-1/b^2-2b-3+4b/b^2-6a+9

$$\frac { b ^ { 2 } - 1 } { b ^ { 2 } - 2 b - 3 } + \frac { 4 b } { b ^ { 2 } - 6 a + 9 }$$

Evaluate

$$\frac{b^{3}-6ab+3b^{2}+6a-3b-9}{\left(b-3\right)\left(9+b^{2}-6a\right)}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{b^{2}-1}{b^{2}-2b-3}.

$$\frac{\left(b-1\right)\left(b+1\right)}{\left(b-3\right)\left(b+1\right)}+\frac{4b}{b^{2}-6a+9}$$

Cancel out b+1 in both numerator and denominator.

$$\frac{b-1}{b-3}+\frac{4b}{b^{2}-6a+9}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of b-3 and b^{2}-6a+9 is \left(b-3\right)\left(-6a+b^{2}+9\right). Multiply \frac{b-1}{b-3} times \frac{-6a+b^{2}+9}{-6a+b^{2}+9}. Multiply \frac{4b}{b^{2}-6a+9} times \frac{b-3}{b-3}.

$$\frac{\left(b-1\right)\left(-6a+b^{2}+9\right)}{\left(b-3\right)\left(-6a+b^{2}+9\right)}+\frac{4b\left(b-3\right)}{\left(b-3\right)\left(-6a+b^{2}+9\right)}$$

Since \frac{\left(b-1\right)\left(-6a+b^{2}+9\right)}{\left(b-3\right)\left(-6a+b^{2}+9\right)} and \frac{4b\left(b-3\right)}{\left(b-3\right)\left(-6a+b^{2}+9\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(b-1\right)\left(-6a+b^{2}+9\right)+4b\left(b-3\right)}{\left(b-3\right)\left(-6a+b^{2}+9\right)}$$

Do the multiplications in \left(b-1\right)\left(-6a+b^{2}+9\right)+4b\left(b-3\right).

$$\frac{-6ba+b^{3}+9b+6a-b^{2}-9+4b^{2}-12b}{\left(b-3\right)\left(-6a+b^{2}+9\right)}$$

Combine like terms in -6ba+b^{3}+9b+6a-b^{2}-9+4b^{2}-12b.

$$\frac{-6ba+b^{3}-3b+6a+3b^{2}-9}{\left(b-3\right)\left(-6a+b^{2}+9\right)}$$

Expand \left(b-3\right)\left(-6a+b^{2}+9\right).

$$\frac{-6ba+b^{3}-3b+6a+3b^{2}-9}{-6ab+18a+b^{3}-3b^{2}+9b-27}$$

Expand

$$\frac{b^{3}-6ab+3b^{2}+6a-3b-9}{\left(b-3\right)\left(9+b^{2}-6a\right)}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{b^{2}-1}{b^{2}-2b-3}.

$$\frac{\left(b-1\right)\left(b+1\right)}{\left(b-3\right)\left(b+1\right)}+\frac{4b}{b^{2}-6a+9}$$

Cancel out b+1 in both numerator and denominator.

$$\frac{b-1}{b-3}+\frac{4b}{b^{2}-6a+9}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of b-3 and b^{2}-6a+9 is \left(b-3\right)\left(-6a+b^{2}+9\right). Multiply \frac{b-1}{b-3} times \frac{-6a+b^{2}+9}{-6a+b^{2}+9}. Multiply \frac{4b}{b^{2}-6a+9} times \frac{b-3}{b-3}.

$$\frac{\left(b-1\right)\left(-6a+b^{2}+9\right)}{\left(b-3\right)\left(-6a+b^{2}+9\right)}+\frac{4b\left(b-3\right)}{\left(b-3\right)\left(-6a+b^{2}+9\right)}$$

Since \frac{\left(b-1\right)\left(-6a+b^{2}+9\right)}{\left(b-3\right)\left(-6a+b^{2}+9\right)} and \frac{4b\left(b-3\right)}{\left(b-3\right)\left(-6a+b^{2}+9\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(b-1\right)\left(-6a+b^{2}+9\right)+4b\left(b-3\right)}{\left(b-3\right)\left(-6a+b^{2}+9\right)}$$

Do the multiplications in \left(b-1\right)\left(-6a+b^{2}+9\right)+4b\left(b-3\right).

$$\frac{-6ba+b^{3}+9b+6a-b^{2}-9+4b^{2}-12b}{\left(b-3\right)\left(-6a+b^{2}+9\right)}$$

Combine like terms in -6ba+b^{3}+9b+6a-b^{2}-9+4b^{2}-12b.

$$\frac{-6ba+b^{3}-3b+6a+3b^{2}-9}{\left(b-3\right)\left(-6a+b^{2}+9\right)}$$

Expand \left(b-3\right)\left(-6a+b^{2}+9\right).

$$\frac{-6ba+b^{3}-3b+6a+3b^{2}-9}{-6ab+18a+b^{3}-3b^{2}+9b-27}$$