<s>[INST] The sum of the coefficients of $x^3$ in $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$ I know how to solve such questions when it's like $(x+y)^n$ but I'm not sure about this one:

In $(1-\frac{x}{2}+\frac{1}{\sqrt x})^8$, What's the sum of the
  coefficients of $x^3$?

 [/INST] The formula you're looking for is
$$
(x+y+z)^8 = \sum_{i+j+k = 8} \begin{pmatrix} 8 \\\ i,j,k \end{pmatrix} x^i y^j z^k
$$
with 
$$
\begin{pmatrix} 8 \\\ i,j,k \end{pmatrix} = \frac{8!}{i!j!k!}.
$$
This is known as the multinomial expansion (it works for more than $3$ variables too, you just have to add more indices and modify the multinomial coefficient accordingly). Using this, then you find when does $x^3$ appear in the expansion
$$
\left( 1 + \left( \frac {-x}2 \right) + \frac 1{\sqrt x} \right)^8 = \sum_{i+j+k = 8} \begin{pmatrix} 8 \\\ i,j,k \end{pmatrix} (-x/2)^j (1/\sqrt{x})^k
$$
In this case we must have $j-(k/2) = 3$, which means $2j-k = 6$, and $0 \le i,j,k \le 8$, or we can rewrite this as $0 \le j \le 8$, $k = 2j-6$ and $i = 8-j-k$. This leaves the cases $(5,3,0)$ and $(2,4,2)$. Computing, the coefficient in front of $x^3$ is
$$
 \begin{pmatrix} 8 \\\ 5,3,0 \end{pmatrix} (-1/2)^3 +  \begin{pmatrix} 8 \\\ 2,4,2 \end{pmatrix} (-1/2)^4 = 77/4.
$$
(I got the $77/4$ using WolframAlpha and/or a calculator, no magic there.)
Hope that helps,
</s>