Solve 3a/a^2-1-6a^2/a^3-1

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

Evaluate

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

Short Solution Steps

Factor a^{2}-1. Factor a^{3}-1.

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

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

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

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

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

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

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

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

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

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

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

Factor

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