Solve c^2+3c+1/c^3-8+c+3/8-c^3

$$\frac { c ^ { 2 } + 3 c + 1 } { c ^ { 3 } - 8 } + \frac { c + 3 } { 8 - c ^ { 3 } }$$

Evaluate

$$\frac{2-2c-c^{2}}{8-c^{3}}$$

Short Solution Steps

Factor c^{3}-8. Factor 8-c^{3}.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(c-2\right)\left(c^{2}+2c+4\right) and \left(c-2\right)\left(-c^{2}-2c-4\right) is \left(c-2\right)\left(c^{2}+2c+4\right). Multiply \frac{c+3}{\left(c-2\right)\left(-c^{2}-2c-4\right)} times \frac{-1}{-1}.

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

Since \frac{c^{2}+3c+1}{\left(c-2\right)\left(c^{2}+2c+4\right)} and \frac{-\left(c+3\right)}{\left(c-2\right)\left(c^{2}+2c+4\right)} have the same denominator, add them by adding their numerators.

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

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

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

Combine like terms in c^{2}+3c+1-c-3.

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

Expand \left(c-2\right)\left(c^{2}+2c+4\right).

$$\frac{c^{2}+2c-2}{c^{3}-8}$$

Expand

$$-\frac{2-2c-c^{2}}{c^{3}-8}$$

Short Solution Steps

Factor c^{3}-8. Factor 8-c^{3}.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(c-2\right)\left(c^{2}+2c+4\right) and \left(c-2\right)\left(-c^{2}-2c-4\right) is \left(c-2\right)\left(c^{2}+2c+4\right). Multiply \frac{c+3}{\left(c-2\right)\left(-c^{2}-2c-4\right)} times \frac{-1}{-1}.

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

Since \frac{c^{2}+3c+1}{\left(c-2\right)\left(c^{2}+2c+4\right)} and \frac{-\left(c+3\right)}{\left(c-2\right)\left(c^{2}+2c+4\right)} have the same denominator, add them by adding their numerators.

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

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

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

Combine like terms in c^{2}+3c+1-c-3.

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

Expand \left(c-2\right)\left(c^{2}+2c+4\right).

$$\frac{c^{2}+2c-2}{c^{3}-8}$$