Solve (1+a/a^{2+1}+2+a/a^{2}-1/a):frac{a^{4}+3a^3+2a}{a^4-1}*(1-1/a^2)^-1

$$( \frac { 1 + a } { a ^ { 2 } + 1 } + \frac { 2 + a } { a ^ { 2 } } - \frac { 1 } { a } ) : \frac { a ^ { 4 } + 3 a ^ { 3 } + 2 a } { a ^ { 4 } - 1 } \cdot ( 1 - \frac { 1 } { a ^ { 2 } } ) ^ { - 1 }$$

Evaluate

$$\frac{1}{a}$$

Short Solution Steps

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

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

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

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

Do the multiplications in \left(1+a\right)a^{2}+\left(2+a\right)\left(a^{2}+1\right).

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

Combine like terms in a^{2}+a^{3}+2a^{2}+2+a^{3}+a.

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

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

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

Since \frac{3a^{2}+2a^{3}+2+a}{a^{2}\left(a^{2}+1\right)} and \frac{a\left(a^{2}+1\right)}{a^{2}\left(a^{2}+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

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

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

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

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

Divide \frac{3a^{2}+a^{3}+2}{a^{2}\left(a^{2}+1\right)} by \frac{a^{4}+3a^{3}+2a}{a^{4}-1} by multiplying \frac{3a^{2}+a^{3}+2}{a^{2}\left(a^{2}+1\right)} by the reciprocal of \frac{a^{4}+3a^{3}+2a}{a^{4}-1}.

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

Factor the expressions that are not already factored in \frac{\left(3a^{2}+a^{3}+2\right)\left(a^{4}-1\right)}{a^{2}\left(a^{2}+1\right)\left(a^{4}+3a^{3}+2a\right)}.

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

Cancel out \left(a^{2}+1\right)\left(a^{3}+3a^{2}+2\right) in both numerator and denominator.

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

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

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

Since \frac{a^{2}}{a^{2}} and \frac{1}{a^{2}} have the same denominator, subtract them by subtracting their numerators.

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

To raise \frac{a^{2}-1}{a^{2}} to a power, raise both numerator and denominator to the power and then divide.

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

Multiply \frac{\left(a-1\right)\left(a+1\right)}{a^{3}} times \frac{\left(a^{2}-1\right)^{-1}}{\left(a^{2}\right)^{-1}} by multiplying numerator times numerator and denominator times denominator.

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

To raise a power to another power, multiply the exponents. Multiply 2 and -1 to get -2.

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

To multiply powers of the same base, add their exponents. Add 3 and -2 to get 1.

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

Calculate a to the power of 1 and get a.

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

Use the distributive property to multiply a-1 by a+1 and combine like terms.

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

Multiply a^{2}-1 and \left(a^{2}-1\right)^{-1} to get 1.

$$\frac{1}{a}$$

Expand

$$\frac{1}{a}$$

Short Solution Steps

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

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

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

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

Do the multiplications in \left(1+a\right)a^{2}+\left(2+a\right)\left(a^{2}+1\right).

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

Combine like terms in a^{2}+a^{3}+2a^{2}+2+a^{3}+a.

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

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

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

Since \frac{3a^{2}+2a^{3}+2+a}{a^{2}\left(a^{2}+1\right)} and \frac{a\left(a^{2}+1\right)}{a^{2}\left(a^{2}+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

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

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

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

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

Divide \frac{3a^{2}+a^{3}+2}{a^{2}\left(a^{2}+1\right)} by \frac{a^{4}+3a^{3}+2a}{a^{4}-1} by multiplying \frac{3a^{2}+a^{3}+2}{a^{2}\left(a^{2}+1\right)} by the reciprocal of \frac{a^{4}+3a^{3}+2a}{a^{4}-1}.

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

Factor the expressions that are not already factored in \frac{\left(3a^{2}+a^{3}+2\right)\left(a^{4}-1\right)}{a^{2}\left(a^{2}+1\right)\left(a^{4}+3a^{3}+2a\right)}.

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

Cancel out \left(a^{2}+1\right)\left(a^{3}+3a^{2}+2\right) in both numerator and denominator.

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

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

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

Since \frac{a^{2}}{a^{2}} and \frac{1}{a^{2}} have the same denominator, subtract them by subtracting their numerators.

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

To raise \frac{a^{2}-1}{a^{2}} to a power, raise both numerator and denominator to the power and then divide.

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

Multiply \frac{\left(a-1\right)\left(a+1\right)}{a^{3}} times \frac{\left(a^{2}-1\right)^{-1}}{\left(a^{2}\right)^{-1}} by multiplying numerator times numerator and denominator times denominator.

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

To raise a power to another power, multiply the exponents. Multiply 2 and -1 to get -2.

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

To multiply powers of the same base, add their exponents. Add 3 and -2 to get 1.

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

Calculate a to the power of 1 and get a.

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

Use the distributive property to multiply a-1 by a+1 and combine like terms.

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

Multiply a^{2}-1 and \left(a^{2}-1\right)^{-1} to get 1.

$$\frac{1}{a}$$