Simplify (x^3-8)/(2-x)

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

Step 1

Simplify the numerator.

Rewrite $8$ as $2^{3}$ .

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

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = x$ and $b = 2$ .

$\frac{(x - 2) (x^{2} + x \cdot 2 + 2^{2})}{2 - x}$

Simplify.

Move $2$ to the left of $x$ .

$\frac{(x - 2) (x^{2} + 2 \cdot x + 2^{2})}{2 - x}$

Raise $2$ to the power of $2$ .

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

Step 2

Simplify terms.

Cancel the common factor of $x - 2$ and $2 - x$ .

Factor $- 1$ out of $x$ .

$\frac{(- 1 (- x) - 2) (x^{2} + 2 x + 4)}{2 - x}$

Rewrite $- 2$ as $- 1 (2)$ .

$\frac{(- 1 (- x) - 1 (2)) (x^{2} + 2 x + 4)}{2 - x}$

Factor $- 1$ out of $- 1 (- x) - 1 (2)$ .

$\frac{- 1 (- x + 2) (x^{2} + 2 x + 4)}{2 - x}$

Reorder terms.

$\frac{- 1 (- x + 2) (x^{2} + 2 x + 4)}{- x + 2}$

Cancel the common factor.

$\frac{- 1 (- x + 2) (x^{2} + 2 x + 4)}{- x + 2}$

Divide $- 1 (x^{2} + 2 x + 4)$ by $1$ .

$- 1 (x^{2} + 2 x + 4)$

Rewrite $- 1 (x^{2} + 2 x + 4)$ as $- (x^{2} + 2 x + 4)$ .

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

Apply the distributive property.

$- x^{2} - (2 x) - 1 \cdot 4$

Step 3

Simplify.

Multiply $2$ by $- 1$ .

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

Multiply $- 1$ by $4$ .

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

Simplify (x^3-8)/(2-x)

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