Divide (x^2+2x+4)÷(x^3-8)

$(x^{2} + 2 x + 4) \div (x^{3} - 8)$

Rewrite the division as a fraction.

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

Simplify the denominator.

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

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

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 $b = 2$ .

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

Simplify.

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

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

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

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

Cancel the common factor of $x^{2} + 2 x + 4$ .

Cancel the common factor.

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

Rewrite the expression.

$\frac{1}{x - 2}$

Divide (x^2+2x+4)÷(x^3-8)

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