Solve a^2-9/2a^2+1*6a^-1/a-3+6a-1/a+3

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

Evaluate

$$\frac{12a^{4}-2a^{3}+12a^{2}+35a+54}{a\left(a+3\right)\left(2a^{2}+1\right)}$$

Short Solution Steps

Multiply \frac{a^{2}-9}{2a^{2}+1} times \frac{6a^{-1}}{a-3} by multiplying numerator times numerator and denominator times denominator.

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

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

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

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

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

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

$$\frac{6a^{2}+18a-54-162\times \frac{1}{a}+12a^{4}+6a^{2}-36a^{3}-18a-2a^{3}-a+6a^{2}+3}{\left(a-3\right)\left(a+3\right)\left(2a^{2}+1\right)}$$

Combine like terms in 6a^{2}+18a-54-162\times \frac{1}{a}+12a^{4}+6a^{2}-36a^{3}-18a-2a^{3}-a+6a^{2}+3.

$$\frac{18a^{2}-a-51-162\times \frac{1}{a}+12a^{4}-38a^{3}}{\left(a-3\right)\left(a+3\right)\left(2a^{2}+1\right)}$$

Expand \left(a-3\right)\left(a+3\right)\left(2a^{2}+1\right).

$$\frac{18a^{2}-a-51-162\times \frac{1}{a}+12a^{4}-38a^{3}}{2a^{4}-17a^{2}-9}$$

Express -162\times \frac{1}{a} as a single fraction.

$$\frac{18a^{2}-a-51+\frac{-162}{a}+12a^{4}-38a^{3}}{2a^{4}-17a^{2}-9}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 18a^{2}-a-51+12a^{4}-38a^{3} times \frac{a}{a}.

$$\frac{\frac{\left(18a^{2}-a-51+12a^{4}-38a^{3}\right)a}{a}+\frac{-162}{a}}{2a^{4}-17a^{2}-9}$$

Since \frac{\left(18a^{2}-a-51+12a^{4}-38a^{3}\right)a}{a} and \frac{-162}{a} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(18a^{2}-a-51+12a^{4}-38a^{3}\right)a-162}{a}}{2a^{4}-17a^{2}-9}$$

Do the multiplications in \left(18a^{2}-a-51+12a^{4}-38a^{3}\right)a-162.

$$\frac{\frac{18a^{3}-a^{2}-51a+12a^{5}-38a^{4}-162}{a}}{2a^{4}-17a^{2}-9}$$

Express \frac{\frac{18a^{3}-a^{2}-51a+12a^{5}-38a^{4}-162}{a}}{2a^{4}-17a^{2}-9} as a single fraction.

$$\frac{18a^{3}-a^{2}-51a+12a^{5}-38a^{4}-162}{a\left(2a^{4}-17a^{2}-9\right)}$$

Factor the expressions that are not already factored.

$$\frac{3\left(a-3\right)\left(4a^{4}-\frac{2}{3}a^{3}+4a^{2}+\frac{35}{3}a+18\right)}{a\left(a-3\right)\left(a+3\right)\left(2a^{2}+1\right)}$$

Cancel out a-3 in both numerator and denominator.

$$\frac{3\left(4a^{4}-\frac{2}{3}a^{3}+4a^{2}+\frac{35}{3}a+18\right)}{a\left(a+3\right)\left(2a^{2}+1\right)}$$

Expand the expression.

$$\frac{12a^{4}-2a^{3}+12a^{2}+35a+54}{2a^{4}+6a^{3}+a^{2}+3a}$$

Expand

$$\frac{12a^{4}-2a^{3}+12a^{2}+35a+54}{a\left(a+3\right)\left(2a^{2}+1\right)}$$

Short Solution Steps

Multiply \frac{a^{2}-9}{2a^{2}+1} times \frac{6a^{-1}}{a-3} by multiplying numerator times numerator and denominator times denominator.

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

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

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

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

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

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

$$\frac{6a^{2}+18a-54-162\times \frac{1}{a}+12a^{4}+6a^{2}-36a^{3}-18a-2a^{3}-a+6a^{2}+3}{\left(a-3\right)\left(a+3\right)\left(2a^{2}+1\right)}$$

Combine like terms in 6a^{2}+18a-54-162\times \frac{1}{a}+12a^{4}+6a^{2}-36a^{3}-18a-2a^{3}-a+6a^{2}+3.

$$\frac{18a^{2}-a-51-162\times \frac{1}{a}+12a^{4}-38a^{3}}{\left(a-3\right)\left(a+3\right)\left(2a^{2}+1\right)}$$

Expand \left(a-3\right)\left(a+3\right)\left(2a^{2}+1\right).

$$\frac{18a^{2}-a-51-162\times \frac{1}{a}+12a^{4}-38a^{3}}{2a^{4}-17a^{2}-9}$$

Express -162\times \frac{1}{a} as a single fraction.

$$\frac{18a^{2}-a-51+\frac{-162}{a}+12a^{4}-38a^{3}}{2a^{4}-17a^{2}-9}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 18a^{2}-a-51+12a^{4}-38a^{3} times \frac{a}{a}.

$$\frac{\frac{\left(18a^{2}-a-51+12a^{4}-38a^{3}\right)a}{a}+\frac{-162}{a}}{2a^{4}-17a^{2}-9}$$

Since \frac{\left(18a^{2}-a-51+12a^{4}-38a^{3}\right)a}{a} and \frac{-162}{a} have the same denominator, add them by adding their numerators.

$$\frac{\frac{\left(18a^{2}-a-51+12a^{4}-38a^{3}\right)a-162}{a}}{2a^{4}-17a^{2}-9}$$

Do the multiplications in \left(18a^{2}-a-51+12a^{4}-38a^{3}\right)a-162.

$$\frac{\frac{18a^{3}-a^{2}-51a+12a^{5}-38a^{4}-162}{a}}{2a^{4}-17a^{2}-9}$$

Express \frac{\frac{18a^{3}-a^{2}-51a+12a^{5}-38a^{4}-162}{a}}{2a^{4}-17a^{2}-9} as a single fraction.

$$\frac{18a^{3}-a^{2}-51a+12a^{5}-38a^{4}-162}{a\left(2a^{4}-17a^{2}-9\right)}$$

Factor the expressions that are not already factored.

$$\frac{3\left(a-3\right)\left(4a^{4}-\frac{2}{3}a^{3}+4a^{2}+\frac{35}{3}a+18\right)}{a\left(a-3\right)\left(a+3\right)\left(2a^{2}+1\right)}$$

Cancel out a-3 in both numerator and denominator.

$$\frac{3\left(4a^{4}-\frac{2}{3}a^{3}+4a^{2}+\frac{35}{3}a+18\right)}{a\left(a+3\right)\left(2a^{2}+1\right)}$$

Expand the expression.

$$\frac{12a^{4}-2a^{3}+12a^{2}+35a+54}{2a^{4}+6a^{3}+a^{2}+3a}$$