Solve x^3-8/x^2+2x+4-x^3+64/x^2-4x+16

$$\frac { x ^ { 3 } - 8 } { x ^ { 2 } + 2 x + 4 } - \frac { x ^ { 3 } + 64 } { x ^ { 2 } - 4 x + 16 }$$

Evaluate

$$-6$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{x^{3}-8}{x^{2}+2x+4}.

$$\frac{\left(x-2\right)\left(x^{2}+2x+4\right)}{x^{2}+2x+4}-\frac{x^{3}+64}{x^{2}-4x+16}$$

Cancel out x^{2}+2x+4 in both numerator and denominator.

$$x-2-\frac{x^{3}+64}{x^{2}-4x+16}$$

Factor the expressions that are not already factored in \frac{x^{3}+64}{x^{2}-4x+16}.

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

Cancel out x^{2}-4x+16 in both numerator and denominator.

$$x-2-\left(x+4\right)$$

To find the opposite of x+4, find the opposite of each term.

$$x-2-x-4$$

Combine x and -x to get 0.

$$-2-4$$

Subtract 4 from -2 to get -6.

$$-6$$

Factor

$$-6$$