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

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

Solve for E

$$E=2$$

$$n\neq -1$$

Steps for Solving Linear Equation

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

$$E=\frac{\left(n+1\right)\left(n^{2}-n+1\right)}{\left(n^{2}-n+1\right)^{2}}-\frac{n^{2}+2n-1-2n^{3}}{n^{3}+1}$$

Cancel out n^{2}-n+1 in both numerator and denominator.

$$E=\frac{n+1}{n^{2}-n+1}-\frac{n^{2}+2n-1-2n^{3}}{n^{3}+1}$$

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

$$E=\frac{n+1}{n^{2}-n+1}-\frac{\left(n-1\right)\left(n+1\right)\left(-2n+1\right)}{\left(n+1\right)\left(n^{2}-n+1\right)}$$

Cancel out n+1 in both numerator and denominator.

$$E=\frac{n+1}{n^{2}-n+1}-\frac{\left(n-1\right)\left(-2n+1\right)}{n^{2}-n+1}$$

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

$$E=\frac{n+1-\left(n-1\right)\left(-2n+1\right)}{n^{2}-n+1}$$

Do the multiplications in n+1-\left(n-1\right)\left(-2n+1\right).

$$E=\frac{n+1+2n^{2}-n-2n+1}{n^{2}-n+1}$$

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

$$E=\frac{-2n+2+2n^{2}}{n^{2}-n+1}$$

Factor the expressions that are not already factored in \frac{-2n+2+2n^{2}}{n^{2}-n+1}.

$$E=\frac{2\left(n^{2}-n+1\right)}{n^{2}-n+1}$$

Cancel out n^{2}-n+1 in both numerator and denominator.

$$E=2$$

Solve for n

$$n\neq -1$$

$$E=2\text{ and }n\neq -1$$