Solve 4a^2-3a+5/a^3-1-1-2a/a^2+a+1+6/1-a

$$\frac { 4 a ^ { 2 } - 3 a + 5 } { a ^ { 3 } - 1 } - \frac { 1 - 2 a } { a ^ { 2 } + a + 1 } + \frac { 6 } { 1 - a }$$

Evaluate

$$\frac{12a}{1-a^{3}}$$

Short Solution Steps

Factor a^{3}-1.

$$\frac{4a^{2}-3a+5}{\left(a-1\right)\left(a^{2}+a+1\right)}-\frac{1-2a}{a^{2}+a+1}+\frac{6}{1-a}$$

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

$$\frac{4a^{2}-3a+5}{\left(a-1\right)\left(a^{2}+a+1\right)}-\frac{\left(1-2a\right)\left(a-1\right)}{\left(a-1\right)\left(a^{2}+a+1\right)}+\frac{6}{1-a}$$

Since \frac{4a^{2}-3a+5}{\left(a-1\right)\left(a^{2}+a+1\right)} and \frac{\left(1-2a\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.

$$\frac{4a^{2}-3a+5-\left(1-2a\right)\left(a-1\right)}{\left(a-1\right)\left(a^{2}+a+1\right)}+\frac{6}{1-a}$$

Do the multiplications in 4a^{2}-3a+5-\left(1-2a\right)\left(a-1\right).

$$\frac{4a^{2}-3a+5+1-a+2a^{2}-2a}{\left(a-1\right)\left(a^{2}+a+1\right)}+\frac{6}{1-a}$$

Combine like terms in 4a^{2}-3a+5+1-a+2a^{2}-2a.

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

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

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

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

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

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

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

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

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

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

$$\frac{-12a}{a^{3}-1}$$

Expand

$$-\frac{12a}{a^{3}-1}$$

Short Solution Steps

Factor a^{3}-1.

$$\frac{4a^{2}-3a+5}{\left(a-1\right)\left(a^{2}+a+1\right)}-\frac{1-2a}{a^{2}+a+1}+\frac{6}{1-a}$$

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

$$\frac{4a^{2}-3a+5}{\left(a-1\right)\left(a^{2}+a+1\right)}-\frac{\left(1-2a\right)\left(a-1\right)}{\left(a-1\right)\left(a^{2}+a+1\right)}+\frac{6}{1-a}$$

Since \frac{4a^{2}-3a+5}{\left(a-1\right)\left(a^{2}+a+1\right)} and \frac{\left(1-2a\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.

$$\frac{4a^{2}-3a+5-\left(1-2a\right)\left(a-1\right)}{\left(a-1\right)\left(a^{2}+a+1\right)}+\frac{6}{1-a}$$

Do the multiplications in 4a^{2}-3a+5-\left(1-2a\right)\left(a-1\right).

$$\frac{4a^{2}-3a+5+1-a+2a^{2}-2a}{\left(a-1\right)\left(a^{2}+a+1\right)}+\frac{6}{1-a}$$

Combine like terms in 4a^{2}-3a+5+1-a+2a^{2}-2a.

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

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

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

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

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

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

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

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

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

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

$$\frac{-12a}{a^{3}-1}$$