Solve x-16/x^3/frac{1{2}-4/x^3}:2x^2+8/x^2+2x+4

$$\frac { x - \frac { 16 } { x ^ { 3 } } } { \frac { 1 } { 2 } - \frac { 4 } { x ^ { 3 } } } : \frac { 2 x ^ { 2 } + 8 } { x ^ { 2 } + 2 x + 4 }$$

Evaluate

$$x+2$$

Short Solution Steps

Divide \frac{x-\frac{16}{x^{3}}}{\frac{1}{2}-\frac{4}{x^{3}}} by \frac{2x^{2}+8}{x^{2}+2x+4} by multiplying \frac{x-\frac{16}{x^{3}}}{\frac{1}{2}-\frac{4}{x^{3}}} by the reciprocal of \frac{2x^{2}+8}{x^{2}+2x+4}.

$$\frac{\left(x-\frac{16}{x^{3}}\right)\left(x^{2}+2x+4\right)}{\left(\frac{1}{2}-\frac{4}{x^{3}}\right)\left(2x^{2}+8\right)}$$

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

$$\frac{\left(\frac{xx^{3}}{x^{3}}-\frac{16}{x^{3}}\right)\left(x^{2}+2x+4\right)}{\left(\frac{1}{2}-\frac{4}{x^{3}}\right)\left(2x^{2}+8\right)}$$

Since \frac{xx^{3}}{x^{3}} and \frac{16}{x^{3}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{xx^{3}-16}{x^{3}}\left(x^{2}+2x+4\right)}{\left(\frac{1}{2}-\frac{4}{x^{3}}\right)\left(2x^{2}+8\right)}$$

Do the multiplications in xx^{3}-16.

$$\frac{\frac{x^{4}-16}{x^{3}}\left(x^{2}+2x+4\right)}{\left(\frac{1}{2}-\frac{4}{x^{3}}\right)\left(2x^{2}+8\right)}$$

Express \frac{x^{4}-16}{x^{3}}\left(x^{2}+2x+4\right) as a single fraction.

$$\frac{\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}}}{\left(\frac{1}{2}-\frac{4}{x^{3}}\right)\left(2x^{2}+8\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and x^{3} is 2x^{3}. Multiply \frac{1}{2} times \frac{x^{3}}{x^{3}}. Multiply \frac{4}{x^{3}} times \frac{2}{2}.

$$\frac{\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}}}{\left(\frac{x^{3}}{2x^{3}}-\frac{4\times 2}{2x^{3}}\right)\left(2x^{2}+8\right)}$$

Since \frac{x^{3}}{2x^{3}} and \frac{4\times 2}{2x^{3}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}}}{\frac{x^{3}-4\times 2}{2x^{3}}\left(2x^{2}+8\right)}$$

Do the multiplications in x^{3}-4\times 2.

$$\frac{\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}}}{\frac{x^{3}-8}{2x^{3}}\left(2x^{2}+8\right)}$$

Express \frac{x^{3}-8}{2x^{3}}\left(2x^{2}+8\right) as a single fraction.

$$\frac{\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}}}{\frac{\left(x^{3}-8\right)\left(2x^{2}+8\right)}{2x^{3}}}$$

Divide \frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}} by \frac{\left(x^{3}-8\right)\left(2x^{2}+8\right)}{2x^{3}} by multiplying \frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}} by the reciprocal of \frac{\left(x^{3}-8\right)\left(2x^{2}+8\right)}{2x^{3}}.

$$\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)\times 2x^{3}}{x^{3}\left(x^{3}-8\right)\left(2x^{2}+8\right)}$$

Cancel out x^{3} in both numerator and denominator.

$$\frac{2\left(x^{2}+2x+4\right)\left(x^{4}-16\right)}{\left(2x^{2}+8\right)\left(x^{3}-8\right)}$$

Factor the expressions that are not already factored.

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

Cancel out 2\left(x-2\right)\left(x^{2}+4\right)\left(x^{2}+2x+4\right) in both numerator and denominator.

$$x+2$$

Expand

$$x+2$$

Short Solution Steps

Divide \frac{x-\frac{16}{x^{3}}}{\frac{1}{2}-\frac{4}{x^{3}}} by \frac{2x^{2}+8}{x^{2}+2x+4} by multiplying \frac{x-\frac{16}{x^{3}}}{\frac{1}{2}-\frac{4}{x^{3}}} by the reciprocal of \frac{2x^{2}+8}{x^{2}+2x+4}.

$$\frac{\left(x-\frac{16}{x^{3}}\right)\left(x^{2}+2x+4\right)}{\left(\frac{1}{2}-\frac{4}{x^{3}}\right)\left(2x^{2}+8\right)}$$

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

$$\frac{\left(\frac{xx^{3}}{x^{3}}-\frac{16}{x^{3}}\right)\left(x^{2}+2x+4\right)}{\left(\frac{1}{2}-\frac{4}{x^{3}}\right)\left(2x^{2}+8\right)}$$

Since \frac{xx^{3}}{x^{3}} and \frac{16}{x^{3}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{xx^{3}-16}{x^{3}}\left(x^{2}+2x+4\right)}{\left(\frac{1}{2}-\frac{4}{x^{3}}\right)\left(2x^{2}+8\right)}$$

Do the multiplications in xx^{3}-16.

$$\frac{\frac{x^{4}-16}{x^{3}}\left(x^{2}+2x+4\right)}{\left(\frac{1}{2}-\frac{4}{x^{3}}\right)\left(2x^{2}+8\right)}$$

Express \frac{x^{4}-16}{x^{3}}\left(x^{2}+2x+4\right) as a single fraction.

$$\frac{\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}}}{\left(\frac{1}{2}-\frac{4}{x^{3}}\right)\left(2x^{2}+8\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and x^{3} is 2x^{3}. Multiply \frac{1}{2} times \frac{x^{3}}{x^{3}}. Multiply \frac{4}{x^{3}} times \frac{2}{2}.

$$\frac{\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}}}{\left(\frac{x^{3}}{2x^{3}}-\frac{4\times 2}{2x^{3}}\right)\left(2x^{2}+8\right)}$$

Since \frac{x^{3}}{2x^{3}} and \frac{4\times 2}{2x^{3}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}}}{\frac{x^{3}-4\times 2}{2x^{3}}\left(2x^{2}+8\right)}$$

Do the multiplications in x^{3}-4\times 2.

$$\frac{\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}}}{\frac{x^{3}-8}{2x^{3}}\left(2x^{2}+8\right)}$$

Express \frac{x^{3}-8}{2x^{3}}\left(2x^{2}+8\right) as a single fraction.

$$\frac{\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}}}{\frac{\left(x^{3}-8\right)\left(2x^{2}+8\right)}{2x^{3}}}$$

Divide \frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}} by \frac{\left(x^{3}-8\right)\left(2x^{2}+8\right)}{2x^{3}} by multiplying \frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)}{x^{3}} by the reciprocal of \frac{\left(x^{3}-8\right)\left(2x^{2}+8\right)}{2x^{3}}.

$$\frac{\left(x^{4}-16\right)\left(x^{2}+2x+4\right)\times 2x^{3}}{x^{3}\left(x^{3}-8\right)\left(2x^{2}+8\right)}$$

Cancel out x^{3} in both numerator and denominator.

$$\frac{2\left(x^{2}+2x+4\right)\left(x^{4}-16\right)}{\left(2x^{2}+8\right)\left(x^{3}-8\right)}$$

Factor the expressions that are not already factored.

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

Cancel out 2\left(x-2\right)\left(x^{2}+4\right)\left(x^{2}+2x+4\right) in both numerator and denominator.

$$x+2$$