Solve 1-(1/a+3+b/a^2-9)diva+3/a^2-6a+9

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

Evaluate

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

Short Solution Steps

Factor a^{2}-9.

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

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

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

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

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

Divide \frac{a-3+b}{\left(a-3\right)\left(a+3\right)} by \frac{a+3}{a^{2}-6a+9} by multiplying \frac{a-3+b}{\left(a-3\right)\left(a+3\right)} by the reciprocal of \frac{a+3}{a^{2}-6a+9}.

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

Factor the expressions that are not already factored in \frac{\left(a-3+b\right)\left(a^{2}-6a+9\right)}{\left(a-3\right)\left(a+3\right)\left(a+3\right)}.

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

Cancel out a-3 in both numerator and denominator.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(a+3\right)^{2}}{\left(a+3\right)^{2}}.

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

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

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

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

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

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

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

Expand \left(a+3\right)^{2}.

$$\frac{12a-ab+3b}{a^{2}+6a+9}$$

Expand

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

Short Solution Steps

Factor a^{2}-9.

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

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

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

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

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

Divide \frac{a-3+b}{\left(a-3\right)\left(a+3\right)} by \frac{a+3}{a^{2}-6a+9} by multiplying \frac{a-3+b}{\left(a-3\right)\left(a+3\right)} by the reciprocal of \frac{a+3}{a^{2}-6a+9}.

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

Factor the expressions that are not already factored in \frac{\left(a-3+b\right)\left(a^{2}-6a+9\right)}{\left(a-3\right)\left(a+3\right)\left(a+3\right)}.

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

Cancel out a-3 in both numerator and denominator.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{\left(a+3\right)^{2}}{\left(a+3\right)^{2}}.

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

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

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

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

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

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

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

Expand \left(a+3\right)^{2}.

$$\frac{12a-ab+3b}{a^{2}+6a+9}$$