Divide (x^3-64)/(x-4)

$\frac{x^{3} - 64}{x - 4}$

Step 1

Simplify the numerator.

Rewrite $64$ as $4^{3}$ .

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

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 $b = 4$ .

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

Simplify.

Move $4$ to the left of $x$ .

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

Raise $4$ to the power of $2$ .

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

Step 2

Cancel the common factor of $x - 4$ .

Cancel the common factor.

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

Divide $x^{2} + 4 x + 16$ by $1$ .

$x^{2} + 4 x + 16$

Divide (x^3-64)/(x-4)

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