Solve 1/4a+4-1/8a-8-1/12a^2+1

$$\frac { 1 } { 4 a + 4 } - \frac { 1 } { 8 a - 8 } - \frac { 1 } { 12 a ^ { 2 } + 1 }$$

Evaluate

$$\frac{12a^{3}-44a^{2}+a+5}{8\left(a^{2}-1\right)\left(12a^{2}+1\right)}$$

Short Solution Steps

Factor 4a+4. Factor 8a-8.

$$\frac{1}{4\left(a+1\right)}-\frac{1}{8\left(a-1\right)}-\frac{1}{12a^{2}+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4\left(a+1\right) and 8\left(a-1\right) is 8\left(a-1\right)\left(a+1\right). Multiply \frac{1}{4\left(a+1\right)} times \frac{2\left(a-1\right)}{2\left(a-1\right)}. Multiply \frac{1}{8\left(a-1\right)} times \frac{a+1}{a+1}.

$$\frac{2\left(a-1\right)}{8\left(a-1\right)\left(a+1\right)}-\frac{a+1}{8\left(a-1\right)\left(a+1\right)}-\frac{1}{12a^{2}+1}$$

Since \frac{2\left(a-1\right)}{8\left(a-1\right)\left(a+1\right)} and \frac{a+1}{8\left(a-1\right)\left(a+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(a-1\right)-\left(a+1\right)}{8\left(a-1\right)\left(a+1\right)}-\frac{1}{12a^{2}+1}$$

Do the multiplications in 2\left(a-1\right)-\left(a+1\right).

$$\frac{2a-2-a-1}{8\left(a-1\right)\left(a+1\right)}-\frac{1}{12a^{2}+1}$$

Combine like terms in 2a-2-a-1.

$$\frac{a-3}{8\left(a-1\right)\left(a+1\right)}-\frac{1}{12a^{2}+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 8\left(a-1\right)\left(a+1\right) and 12a^{2}+1 is 8\left(a-1\right)\left(a+1\right)\left(12a^{2}+1\right). Multiply \frac{a-3}{8\left(a-1\right)\left(a+1\right)} times \frac{12a^{2}+1}{12a^{2}+1}. Multiply \frac{1}{12a^{2}+1} times \frac{8\left(a-1\right)\left(a+1\right)}{8\left(a-1\right)\left(a+1\right)}.

$$\frac{\left(a-3\right)\left(12a^{2}+1\right)}{8\left(a-1\right)\left(a+1\right)\left(12a^{2}+1\right)}-\frac{8\left(a-1\right)\left(a+1\right)}{8\left(a-1\right)\left(a+1\right)\left(12a^{2}+1\right)}$$

Since \frac{\left(a-3\right)\left(12a^{2}+1\right)}{8\left(a-1\right)\left(a+1\right)\left(12a^{2}+1\right)} and \frac{8\left(a-1\right)\left(a+1\right)}{8\left(a-1\right)\left(a+1\right)\left(12a^{2}+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(a-3\right)\left(12a^{2}+1\right)-8\left(a-1\right)\left(a+1\right)}{8\left(a-1\right)\left(a+1\right)\left(12a^{2}+1\right)}$$

Do the multiplications in \left(a-3\right)\left(12a^{2}+1\right)-8\left(a-1\right)\left(a+1\right).

$$\frac{12a^{3}+a-36a^{2}-3-8a^{2}-8a+8a+8}{8\left(a-1\right)\left(a+1\right)\left(12a^{2}+1\right)}$$

Combine like terms in 12a^{3}+a-36a^{2}-3-8a^{2}-8a+8a+8.

$$\frac{12a^{3}+a-44a^{2}+5}{8\left(a-1\right)\left(a+1\right)\left(12a^{2}+1\right)}$$

Expand 8\left(a-1\right)\left(a+1\right)\left(12a^{2}+1\right).

$$\frac{12a^{3}+a-44a^{2}+5}{96a^{4}-88a^{2}-8}$$

Factor

$$\frac{12a^{3}-44a^{2}+a+5}{8\left(a^{2}-1\right)\left(12a^{2}+1\right)}$$