Solve (1/x^2+3x+1/x^2-3x):x^2+x+1/x^2-9

$$( \frac { 1 } { x ^ { 2 } + 3 x } + \frac { 1 } { x ^ { 2 } - 3 x } ) : \frac { x ^ { 2 } + x + 1 } { x ^ { 2 } - 9 }$$

Evaluate

$$\frac{2}{x^{2}+x+1}$$

Short Solution Steps

Factor x^{2}+3x. Factor x^{2}-3x.

$$\frac{\frac{1}{x\left(x+3\right)}+\frac{1}{x\left(x-3\right)}}{\frac{x^{2}+x+1}{x^{2}-9}}$$

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

$$\frac{\frac{x-3}{x\left(x-3\right)\left(x+3\right)}+\frac{x+3}{x\left(x-3\right)\left(x+3\right)}}{\frac{x^{2}+x+1}{x^{2}-9}}$$

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

$$\frac{\frac{x-3+x+3}{x\left(x-3\right)\left(x+3\right)}}{\frac{x^{2}+x+1}{x^{2}-9}}$$

Combine like terms in x-3+x+3.

$$\frac{\frac{2x}{x\left(x-3\right)\left(x+3\right)}}{\frac{x^{2}+x+1}{x^{2}-9}}$$

Cancel out x in both numerator and denominator.

$$\frac{\frac{2}{\left(x-3\right)\left(x+3\right)}}{\frac{x^{2}+x+1}{x^{2}-9}}$$

Divide \frac{2}{\left(x-3\right)\left(x+3\right)} by \frac{x^{2}+x+1}{x^{2}-9} by multiplying \frac{2}{\left(x-3\right)\left(x+3\right)} by the reciprocal of \frac{x^{2}+x+1}{x^{2}-9}.

$$\frac{2\left(x^{2}-9\right)}{\left(x-3\right)\left(x+3\right)\left(x^{2}+x+1\right)}$$

Factor the expressions that are not already factored.

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

Cancel out \left(x-3\right)\left(x+3\right) in both numerator and denominator.

$$\frac{2}{x^{2}+x+1}$$

Expand

$$\frac{2}{x^{2}+x+1}$$

Short Solution Steps

Factor x^{2}+3x. Factor x^{2}-3x.

$$\frac{\frac{1}{x\left(x+3\right)}+\frac{1}{x\left(x-3\right)}}{\frac{x^{2}+x+1}{x^{2}-9}}$$

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

$$\frac{\frac{x-3}{x\left(x-3\right)\left(x+3\right)}+\frac{x+3}{x\left(x-3\right)\left(x+3\right)}}{\frac{x^{2}+x+1}{x^{2}-9}}$$

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

$$\frac{\frac{x-3+x+3}{x\left(x-3\right)\left(x+3\right)}}{\frac{x^{2}+x+1}{x^{2}-9}}$$

Combine like terms in x-3+x+3.

$$\frac{\frac{2x}{x\left(x-3\right)\left(x+3\right)}}{\frac{x^{2}+x+1}{x^{2}-9}}$$

Cancel out x in both numerator and denominator.

$$\frac{\frac{2}{\left(x-3\right)\left(x+3\right)}}{\frac{x^{2}+x+1}{x^{2}-9}}$$

Divide \frac{2}{\left(x-3\right)\left(x+3\right)} by \frac{x^{2}+x+1}{x^{2}-9} by multiplying \frac{2}{\left(x-3\right)\left(x+3\right)} by the reciprocal of \frac{x^{2}+x+1}{x^{2}-9}.

$$\frac{2\left(x^{2}-9\right)}{\left(x-3\right)\left(x+3\right)\left(x^{2}+x+1\right)}$$

Factor the expressions that are not already factored.

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

Cancel out \left(x-3\right)\left(x+3\right) in both numerator and denominator.

$$\frac{2}{x^{2}+x+1}$$