Solve f(x)=x^3-1/x^2+1/x^4+frac{1{x^2+1}}

$$f ( x ) = \frac { x ^ { 3 } - \frac { 1 } { x ^ { 2 } + 1 } } { x ^ { 4 } + \frac { 1 } { x ^ { 2 } + 1 } }$$

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{3} times \frac{x^{2}+1}{x^{2}+1}.

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

Since \frac{x^{3}\left(x^{2}+1\right)}{x^{2}+1} and \frac{1}{x^{2}+1} have the same denominator, subtract them by subtracting their numerators.

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{4} times \frac{x^{2}+1}{x^{2}+1}.

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

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

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

Do the multiplications in x^{4}\left(x^{2}+1\right)+1.

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

Divide \frac{x^{5}+x^{3}-1}{x^{2}+1} by \frac{x^{6}+x^{4}+1}{x^{2}+1} by multiplying \frac{x^{5}+x^{3}-1}{x^{2}+1} by the reciprocal of \frac{x^{6}+x^{4}+1}{x^{2}+1}.

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

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

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

Expand

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{3} times \frac{x^{2}+1}{x^{2}+1}.

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

Since \frac{x^{3}\left(x^{2}+1\right)}{x^{2}+1} and \frac{1}{x^{2}+1} have the same denominator, subtract them by subtracting their numerators.

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

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

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x^{4} times \frac{x^{2}+1}{x^{2}+1}.

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

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

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

Do the multiplications in x^{4}\left(x^{2}+1\right)+1.

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

Divide \frac{x^{5}+x^{3}-1}{x^{2}+1} by \frac{x^{6}+x^{4}+1}{x^{2}+1} by multiplying \frac{x^{5}+x^{3}-1}{x^{2}+1} by the reciprocal of \frac{x^{6}+x^{4}+1}{x^{2}+1}.

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

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

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