Divide (x^3+729)/(x+9)

$\frac{x^{3} + 729}{x + 9}$

Step 1

Simplify the numerator.

Rewrite $729$ as $9^{3}$ .

$\frac{x^{3} + 9^{3}}{x + 9}$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 9$ .

$\frac{(x + 9) (x^{2} - x \cdot 9 + 9^{2})}{x + 9}$

Simplify.

Multiply $9$ by $- 1$ .

$\frac{(x + 9) (x^{2} - 9 x + 9^{2})}{x + 9}$

Raise $9$ to the power of $2$ .

$\frac{(x + 9) (x^{2} - 9 x + 81)}{x + 9}$

Step 2

Cancel the common factor of $x + 9$ .

Cancel the common factor.

$\frac{(x + 9) (x^{2} - 9 x + 81)}{x + 9}$

Divide $x^{2} - 9 x + 81$ by $1$ .

$x^{2} - 9 x + 81$

Divide (x^3+729)/(x+9)

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