Solve (1/b^2-9-1/b^2+9)(1-9/b^2)(1+b^2/9)

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

Evaluate

$$\frac{2}{b^{2}}$$

Short Solution Steps

Factor b^{2}-9.

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

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

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

Since \frac{b^{2}+9}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)} and \frac{\left(b-3\right)\left(b+3\right)}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)} have the same denominator, subtract them by subtracting their numerators.

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

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

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

Combine like terms in b^{2}+9-b^{2}-3b+3b+9.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{b^{2}}{b^{2}}.

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

Since \frac{b^{2}}{b^{2}} and \frac{9}{b^{2}} have the same denominator, subtract them by subtracting their numerators.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{9}{9}.

$$\frac{18}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)}\times \frac{b^{2}-9}{b^{2}}\left(\frac{9}{9}+\frac{b^{2}}{9}\right)$$

Since \frac{9}{9} and \frac{b^{2}}{9} have the same denominator, add them by adding their numerators.

$$\frac{18}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)}\times \frac{b^{2}-9}{b^{2}}\times \frac{9+b^{2}}{9}$$

Multiply \frac{18}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)} times \frac{b^{2}-9}{b^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{18\left(b^{2}-9\right)}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)b^{2}}\times \frac{9+b^{2}}{9}$$

Multiply \frac{18\left(b^{2}-9\right)}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)b^{2}} times \frac{9+b^{2}}{9} by multiplying numerator times numerator and denominator times denominator.

$$\frac{18\left(b^{2}-9\right)\left(9+b^{2}\right)}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)b^{2}\times 9}$$

Cancel out 9\left(b^{2}+9\right) in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Cancel out \left(b-3\right)\left(b+3\right) in both numerator and denominator.

$$\frac{2}{b^{2}}$$

Expand

$$\frac{2}{b^{2}}$$

Short Solution Steps

Factor b^{2}-9.

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

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

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

Since \frac{b^{2}+9}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)} and \frac{\left(b-3\right)\left(b+3\right)}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)} have the same denominator, subtract them by subtracting their numerators.

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

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

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

Combine like terms in b^{2}+9-b^{2}-3b+3b+9.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{b^{2}}{b^{2}}.

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

Since \frac{b^{2}}{b^{2}} and \frac{9}{b^{2}} have the same denominator, subtract them by subtracting their numerators.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{9}{9}.

$$\frac{18}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)}\times \frac{b^{2}-9}{b^{2}}\left(\frac{9}{9}+\frac{b^{2}}{9}\right)$$

Since \frac{9}{9} and \frac{b^{2}}{9} have the same denominator, add them by adding their numerators.

$$\frac{18}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)}\times \frac{b^{2}-9}{b^{2}}\times \frac{9+b^{2}}{9}$$

Multiply \frac{18}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)} times \frac{b^{2}-9}{b^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{18\left(b^{2}-9\right)}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)b^{2}}\times \frac{9+b^{2}}{9}$$

Multiply \frac{18\left(b^{2}-9\right)}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)b^{2}} times \frac{9+b^{2}}{9} by multiplying numerator times numerator and denominator times denominator.

$$\frac{18\left(b^{2}-9\right)\left(9+b^{2}\right)}{\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)b^{2}\times 9}$$

Cancel out 9\left(b^{2}+9\right) in both numerator and denominator.

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

Factor the expressions that are not already factored.

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

Cancel out \left(b-3\right)\left(b+3\right) in both numerator and denominator.

$$\frac{2}{b^{2}}$$