Solve (1/8-x^3):(x-1/2)

$$( \frac { 1 } { 8 } - x ^ { 3 } ) : ( x - \frac { 1 } { 2 } )$$

Evaluate

$$-x^{2}-\frac{x}{2}-\frac{1}{4}$$

Short Solution Steps

Factor the expressions that are not already factored.

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

Cancel out 2x-1 in both numerator and denominator.

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

Expand the expression.

$$\frac{-\frac{1}{2}x^{2}-\frac{1}{4}x-\frac{1}{8}}{\frac{1}{2}}$$

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

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

Use the distributive property to multiply -\frac{1}{2}x^{2}-\frac{1}{4}x-\frac{1}{8} by 2.

$$-x^{2}-\frac{1}{2}x-\frac{1}{4}$$

Expand

$$-x^{2}-\frac{x}{2}-\frac{1}{4}$$

Short Solution Steps

Factor the expressions that are not already factored.

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

Cancel out 2x-1 in both numerator and denominator.

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

Expand the expression.

$$\frac{-\frac{1}{2}x^{2}-\frac{1}{4}x-\frac{1}{8}}{\frac{1}{2}}$$

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

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

Use the distributive property to multiply -\frac{1}{2}x^{2}-\frac{1}{4}x-\frac{1}{8} by 2.

$$-x^{2}-\frac{1}{2}x-\frac{1}{4}$$