Solve frac{1+z+z^{2}}{z^{3}+z^2+z+1}-z^2-z+1/z^4+2z^2+1=1/z+1

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

Solve for z

$$z=1$$

Steps for Solving Linear Equation

Variable z cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by \left(z+1\right)\left(z^{2}+1\right)^{2}, the least common multiple of z^{3}+z^{2}+z+1,z^{4}+2z^{2}+1,z+1.

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

Use the distributive property to multiply z^{2}+1 by 1+z+z^{2} and combine like terms.

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

Use the distributive property to multiply z+1 by z^{2}-z+1 and combine like terms.

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

To find the opposite of z^{3}+1, find the opposite of each term.

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

Combine z^{3} and -z^{3} to get 0.

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

Subtract 1 from 1 to get 0.

$$2z^{2}+z^{4}+z=\left(z^{2}+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(z^{2}+1\right)^{2}.

$$2z^{2}+z^{4}+z=\left(z^{2}\right)^{2}+2z^{2}+1$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$2z^{2}+z^{4}+z=z^{4}+2z^{2}+1$$

Subtract z^{4} from both sides.

$$2z^{2}+z^{4}+z-z^{4}=2z^{2}+1$$

Combine z^{4} and -z^{4} to get 0.

$$2z^{2}+z=2z^{2}+1$$

Subtract 2z^{2} from both sides.

$$2z^{2}+z-2z^{2}=1$$

Combine 2z^{2} and -2z^{2} to get 0.

$$z=1$$