Solve x^3+8/x+2-x^3+2^3/x+2

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

Evaluate

$$0$$

Short Solution Steps

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

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

Cancel out x+2 in both numerator and denominator.

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

Calculate 2 to the power of 3 and get 8.

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

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

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

Cancel out x+2 in both numerator and denominator.

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

To find the opposite of x^{2}-2x+4, find the opposite of each term.

$$x^{2}-2x+4-x^{2}+2x-4$$

Combine x^{2} and -x^{2} to get 0.

$$-2x+4+2x-4$$

Combine -2x and 2x to get 0.

$$4-4$$

Subtract 4 from 4 to get 0.

$$0$$

Factor

$$0$$