<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] $$
\begin{align}
\left(\frac1{\sqrt{x}}+1-\frac12x\right)^8
&=\sum_{k=0}^8\binom{8}{k}\left(\frac1{\sqrt{x}}\right)^k\left(1-\frac12x\right)^{8-k}\\
&=\color{#00A000}{\sum_{k=0}^4\binom{8}{2k}\left(\frac1x\right)^k\left(1-\frac12x\right)^{8-2k}}\\
&+\color{#C00000}{\frac1{\sqrt{x}}\sum_{k=0}^3\binom{8}{2k+1}\left(\frac1x\right)^k\left(1-\frac12x\right)^{7-2k}}\tag{1}
\end{align}
$$
The terms from the red part of $(1)$ will all have fractional powers of $x$ in them, so we only need to consider the green part of $(1)$. Furthermore, for each $k$, the highest power of $x$ in the green part is $8-3k$. Therefore, we only need to consider $k=0$ and $k=1$.
For $k=0$, the term in the green sum is $\left(1-\frac12x\right)^8$ and the coefficient of $x^3$ there is $\binom{8}{3}\left(-\frac12\right)^3$.
For $k=1$, the term in the green sum is $\binom{8}{2}\frac1x\left(1-\frac12x\right)^6$ and the coefficient of $x^3$ there is $\binom{8}{2}$ times the coefficient of $x^4$ in $\left(1-\frac12x\right)^6$, which is $\binom{8}{2}\binom{6}{4}\left(-\frac12\right)^4$.
Therefore, the coefficient of $x^3$ in $\left(\frac1{\sqrt{x}}+1-\frac12x\right)^8$ is $\binom{8}{3}\left(-\frac12\right)^3+\binom{8}{2}\binom{6}{4}\left(-\frac12\right)^4=\frac{77}{4}$.
</s>