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

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

Step 1

Simplify the numerator.

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

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

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 $b = 2$ .

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

Simplify.

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

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

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

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

Step 2

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

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

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

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

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