Are there expansions of the expression $(a+b)^{1/n}$? Is there an expansion of the expression in the bracket such as
$$ \sqrt{a + b} = (a + b)^{1/2}$$
If not do you know of a method that lets us solve such expression and ones with higher roots?
 A: never underestimate the binomial theorem and the coefficients associated with it! they have played a pivotal role in the development of mathematical analysis.
whilst Pascal's triangle well deserves the attention it receives, the development for indices other than a non-negative integer opens new doors. 
it is worth remembering that the expression $(1+x)^y$ is a function of both $x$ and $y$, although we usually focus on $x$ as the variable of interest. 
here is one way of writing the main assertion of the binomial theorem which treats the variables on a more equal footing. as we can see, this leads immediately to a moderately interesting combinatorial result concerning the reciprocals of certain sets of integers:
for any real $a$ and complex number $z$ with $|z| \lt 1$ we have:
$$
(1-z)^a = \sum_{n=0}^{\infty}\sum_{k=1}^n \frac{(-1)^k s(n,k)}{n!}a^kz^n \tag{1}
$$
where the symbol $s(n,k)$ is the unsigned Stirling number of the first kind
since the left-hand side can be developed as:
$$
e^{a\log(1-z)} = \sum_{k=0}^{\infty} \frac{(a\log(1-z))^k}{k!} \tag{2}
$$
we may, by comparing coefficients of $a^k$ in (1) and (2), deduce that:
$$
(\log(1-z))^k = (-1)^k k! \sum_{n=k}^{\infty} s(n,k)\frac{z^n}{n!} \tag{3}
$$
now, since we know that:
$$
\log(1-z) = -\sum_{m=1}^{\infty} \frac{z^m}m
$$
the term involving $z^n$ in $(\log(1-z))^k$ may also be evaluated as the sum:
$$
\sum_{m_1+\dots+m_k=n} \frac1{m_1m_2\dots m_k}
$$
where each $m_i \gt 0$
as a worked example, take $n=7$ and $k=3$ so that the above sum is:
$$
3\frac1{1 \cdot 1 \cdot5}+6\frac1{1 \cdot 2 \cdot 4}+3\frac1{1 \cdot 3 \cdot 3}+3\frac1{2 \cdot 2 \cdot 3} = \frac{29}{15}
$$
which is given by the term from (3) as:
$$
\frac{3!}{7!}s(7,3) = \frac6{5040}\cdot 1624 = \frac{29}{15}
$$
