We have, $\sqrt{3+x}+\sqrt{5+x}$
$\begin{aligned}
& \text { Here, } 3 < x < 5 \\
& \therefore \quad x^{1 / 2}\left(1+\frac{3}{x}\right)^{1 / 2}+5^{1 / 2}\left(1+\frac{x}{5}\right)^{1 / 2} \\
& x^{1 / 2}\left[1+\frac{1}{2}\left(\frac{3}{x}\right)+\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{9}{x^2}\right) \cdots\right] \\
& +5^{1 / 2}\left[1+\frac{1}{2}\left(\frac{x}{5}\right)+\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{x^2}{5^2}\right)+\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)\left(\frac{x^3}{5^3}\right)\right]
\end{aligned}$
$\therefore$ Coefficient of $x^{-3 / 2}$ is,
$\frac{1}{2}\left(\frac{1}{2}-1\right) 9=\frac{1}{2} \times-\frac{1}{2} \times 9=-\frac{9}{4}$
and coefficient of $x^3$ is,
$\begin{aligned}
& 5^{1 / 2} \times \frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right) \frac{1}{5^3} \\
& =5^{\frac{1}{2}-3} \times \frac{1}{2} \times-\frac{1}{2} \times-\frac{3}{2}=5^{\frac{-5}{2}} \times \frac{3}{8}
\end{aligned}$
Sum of coefficient of $x^{-3 / 2}$ and
$x^3=\frac{-9}{4}+\frac{3}{8} 5^{-5 / 2}=\frac{-18+3(5)^{-5 / 2}}{8}$
$\left({ }^*\right)$ No option is matched.