Solve +1+a/a^2-1+a-4/a-1*(a^2-4)

$$\frac { + 1 + a } { a ^ { 2 } - 1 } + \frac { a - 4 } { a - 1 } * ( a ^ { 2 } - 4 )$$

Evaluate

$$\frac{a^{3}-4a^{2}-4a+17}{a-1}$$

Short Solution Steps

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

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

Cancel out a+1 in both numerator and denominator.

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

Express \frac{a-4}{a-1}\left(a^{2}-4\right) as a single fraction.

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

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

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

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

$$\frac{1+a^{3}-4a-4a^{2}+16}{a-1}$$

Combine like terms in 1+a^{3}-4a-4a^{2}+16.

$$\frac{17+a^{3}-4a-4a^{2}}{a-1}$$

Expand

$$\frac{a^{3}-4a^{2}-4a+17}{a-1}$$

Short Solution Steps

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

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

Cancel out a+1 in both numerator and denominator.

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

Express \frac{a-4}{a-1}\left(a^{2}-4\right) as a single fraction.

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

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

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

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

$$\frac{1+a^{3}-4a-4a^{2}+16}{a-1}$$

Combine like terms in 1+a^{3}-4a-4a^{2}+16.

$$\frac{17+a^{3}-4a-4a^{2}}{a-1}$$