# How do you simplify (x-2)/(x^3-8)?

May 12, 2018

$\frac{1}{{x}^{2} + 2 x + 4}$

#### Explanation:

${x}^{3} - 8 = \left(x - 2\right) \left({x}^{2} + 2 x + 4\right)$
The above uses the rule ${x}^{3} - {y}^{3} = \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)$

$\frac{x - 2}{{x}^{3} - 8} = \frac{x - 2}{\left(x - 2\right) \left({x}^{2} + 2 x + 4\right)} = \frac{1}{{x}^{2} + 2 x + 4}$

May 12, 2018

$\frac{1}{{x}^{2} + 2 x + 4} , x \ne 2$

#### Explanation:

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

In rational expressions, any variable on the denominator must not equal zero, so we can find restrictions from the original equation initially to save time in the end:
$\text{ } {x}^{3} - 8 \ne 0$
$\text{ } {x}^{3} \ne 8$
$\text{ } x \ne 2$

Continue solving by putting $8$ into exponent form.
$= \frac{x - 2}{{x}^{3} - {2}^{3}}$

Factor the difference of cubes (${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$).
$= \frac{x - 2}{\left(x - 2\right) \left({x}^{2} + 2 x + 4\right)}$

Reduce the fraction with $x - 2$.
$= \frac{\cancel{x - 2}}{\cancel{x - 2} \left({x}^{2} + 2 x + 4\right)}$
$= \frac{1}{{x}^{2} + 2 x + 4} , x \ne 2$