# What are the asymptotes for  g(x)= (x^3+1)/(x+1)?

Jul 13, 2015

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

with exclusion $x \ne - 1$

The polynomial ${x}^{2} - x + 1$ has no asymptotes, so neither has $g \left(x\right)$

#### Explanation:

The sum of cubes identity is:

${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$

So

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

So

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

with exclusion $x \ne - 1$

The polynomial ${x}^{2} - x + 1$ has no linear asymptotes, so neither has $g \left(x\right)$.