Solve z+1/z^2+z+1+z^4-1/z^2-z^3=

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

Evaluate

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{z^{4}-1}{z^{2}-z^{3}}.

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

Extract the negative sign in -1+z.

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

Cancel out -z+1 in both numerator and denominator.

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

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

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

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

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

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

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

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

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

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

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

Expand

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

Short Solution Steps

Factor the expressions that are not already factored in \frac{z^{4}-1}{z^{2}-z^{3}}.

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

Extract the negative sign in -1+z.

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

Cancel out -z+1 in both numerator and denominator.

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

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

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

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

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

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

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

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

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

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

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