So far I got it down to ((4/3)n^3+(11/6)n^2)/((4/3)n^3)+(2n^2)+(2/3)n).
Would I be able to cancel out the ((4/3)n^3)s? So that it would be ((11/6)n^2)/((2n^2)+((2/3)n))?
Or can I simplify that even futher by combining the (n^2)s?
To determine whether you can cancel out the terms, let's factor the numerator and denominator individually first.
For the numerator ((4/3)n^3 + (11/6)n^2), we can factor out an (n^2):
4/3 * n^3 + 11/6 * n^2
= (4/3 * n) * n^2 + (11/6) * n^2
= (4/3n + 11/6) * n^2
And for the denominator ((4/3)n^3 + 2n^2 + (2/3)n), we can factor out an n:
(4/3)n^3 + 2n^2 + (2/3)n
= n * (4/3 * n^2 + 2n + 2/3)
Now, let's compare the simplified forms of the numerator and denominator:
Numerator: (4/3n + 11/6) * n^2
Denominator: n * (4/3 * n^2 + 2n + 2/3)
As you can see, the n^2 terms in the numerator and denominator cannot be canceled out because there is an additional factor of n in the denominator. Therefore, you cannot simplify the expression further by canceling out the ((4/3)n^3).
However, you can simplify the expression by multiplying through by the reciprocal of the denominator to get rid of the fraction in the denominator:
((4/3n + 11/6) * n^2) / (n * (4/3 * n^2 + 2n + 2/3))
= ((4/3n + 11/6) * n^2) * (1 / (n * (4/3 * n^2 + 2n + 2/3)))
= (4/3n + 11/6) * (n^2 / (n * (4/3 * n^2 + 2n + 2/3)))
= (4/3n + 11/6) * (1 / ((4/3 * n + 2) * n + 2/3))
So, the simplified expression would be ((4/3n + 11/6) * (1 / ((4/3 * n + 2) * n + 2/3))).