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

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Answer 1 · 2015-07-13T20:28:08

#g(x) = (x^3+1)/(x+1) = ((x+1)(x^2-x+1))/(x+1) = x^2-x+1#

with exclusion #x != -1#

The polynomial #x^2-x+1# has no asymptotes, so neither has #g(x)#

The sum of cubes identity is:

#a^3+b^3 = (a+b)(a^2-ab+b^2)#

So

#x^3+1 = x^3+1^3 = (x+1)(x^2-x+1)#

So

#g(x) = (x^3+1)/(x+1) = ((x+1)(x^2-x+1))/(x+1) = x^2-x+1#

with exclusion #x != -1#

The polynomial #x^2-x+1# has no linear asymptotes, so neither has #g(x)#.