Note that the integral isn't defined at $x=2$ but we can compute the principal value as so:
Define $$f(z)=\frac{e^{i z}}{(z-2)(z-2i)(z+2i)}$$ so that $$\operatorname{PV}\int_{-\infty}^{\infty} \frac{\sin(x)}{x^3-2x^2+4x-8}\,dx=\Im \left( \operatorname{PV}\int_{-\infty}^{\infty} f(x) \, dx\right)$$
By Cauchy's residue theorem
\begin{align*}\oint_{C} f(z) \, dz &= 2\pi i \operatorname{Res} \Big[f(z),z=2i\Big]=2\pi i \lim_{z\to 2i}\left(\frac{e^{iz}}{(z-2)(z+2i)}\right)=-\frac{(1+i)\pi}{8e^2}\\
&= \int_{\psi_1} f(x)\,dx+\int_{\gamma_\varepsilon}f(z) \, dz + \int_{\psi_2} f(x) \, dx + \int_{\Gamma_R} f(z) \,dz\end{align*}
By Jordan's lemma, it can be shown that the integral over $\Gamma_R$ vanishes as $R \to \infty$ so we have that $$\int_{-\infty}^{2-\varepsilon} f(x) \, dx+\int_{2+\varepsilon}^{\infty} f(x) \, dx + \int_{\gamma_\varepsilon} f(z)\,dz=-\frac{(1+i)\pi}{8e^2}$$
We will now parametrise over $\gamma_{\varepsilon}$ by setting $z=2+\varepsilon e^{i \theta} \implies dz = i \varepsilon e^{i \theta} \, d\theta$ to get
$$\int_{\gamma_{\varepsilon}} f(z) \, dz=\int_{\pi}^{0}\frac{ie^{i\left(2+\varepsilon e^{i\theta}\right)}}{\left(\varepsilon e^{i\theta}+(2-2i)\right)\left(\varepsilon e^{i\theta}+(2+2i)\right)}\, d \theta$$
Now taking the limit as $\varepsilon \to 0^+$ we have $$\operatorname{PV}\int_{-\infty}^{\infty} f(x) \, dx = -\frac{(1+i)\pi}{8e^2} - \lim_{\varepsilon \to 0^+} \int_{\pi}^{0}\frac{ie^{i\left(2+\varepsilon e^{i\theta}\right)}}{\left(\varepsilon e^{i\theta}+(2-2i)\right)\left(\varepsilon e^{i\theta}+(2+2i)\right)}\, d \theta$$
It can be shown that $f_n \to f$ uniformly and we are on a finite interval so we can bring the limit inside to get
$$\operatorname{PV}\int_{-\infty}^{\infty} f(x) \, dx = -\frac{(1+i)\pi}{8e^2} - \int_{\pi}^{0} \frac{i}{8} e^{2i} \,d\theta=-\frac{(1+i)\pi}{8e^2}+\frac{\pi i}{8} e^{2i}$$
Taking the imaginary part, we can now conclude $$\boxed{\operatorname{PV}\int_{-\infty}^{\infty} \frac{\sin(x)}{x^3-2x^2+4x-8}\,dx=-\frac{\pi}{8e^2}+\frac{\pi}{8} \cos (2)}$$
