Given, $\int x^3 \cdot e^{2 x} d x=\frac{e^{2 x}}{8} f(x)+c$

By applying integration by parts, we get
$$
\begin{aligned}
& \left.x^3 \cdot \int e^{2 x} d x-\int\left(\frac{d}{d x} x^3 \int e^{2 x} d x\right) d x\right)=\frac{e^{2 x}}{8}+f(x)+c \\
& x^3 \cdot \frac{e^{2 x}}{2}-\int 3 x^2 \cdot \frac{e^{2 x}}{2} d x=\frac{e^{2 x}}{8} f(x)+c \\
& \frac{1}{2} x^3 e^{2 x}-\frac{3}{2}\left[x^2 \cdot \frac{e^{2 x}}{2}-\int 2 x \cdot \frac{e^{2 x}}{2} d x\right]=\frac{e^{2 x}}{8} f(x)+c
\end{aligned}
$$
[Again applying integration by parts]
$$
\begin{array}{r}
\frac{1}{2} x^3 e^{2 x}-\frac{3}{4} x^2 \cdot e^{2 x}+\frac{3}{2} \int x \cdot e^{2 x} d x=\frac{e^{2 x}}{8} f(x)+c \\
\frac{1}{2} x^3 e^{2 x}-\frac{3}{4} x^2 \cdot e^{2 x}+\frac{3}{2}\left[x \frac{e^{2 x}}{2}-\int 1 \cdot \frac{e^{2 x}}{2} d x\right] \\
=\frac{e^{2 x}}{8} f(x)+c
\end{array}
$$
$$
\begin{aligned}
& \frac{1}{2} x^3 e^{2 x}-\frac{3}{4} x^2 e^{2 x}+\frac{3}{4} x e^{2 x}-\frac{3}{4} \cdot \frac{e^{2 x}}{2}=\frac{e^{2 x}}{8} f(x)+c \\
& \frac{e^{2 x}}{8}\left[4 x^3-6 x^2+6 x-3\right]+c_1=\frac{e^{2 x}}{8} f(x)+c
\end{aligned}
$$
$\therefore$ On comparison, we get
$$
f(x)=4 x^3-6 x^2+6 x-3
$$

But given,
$$
\begin{aligned}
& f(x)=1 \\
& 4 x^3-6 x^2+6 x-3=1 \\
& 4 x^3-6 x^2+6 x-4=0
\end{aligned}
$$

Divided by 2, we get
$$
\begin{array}{lr}
& 2 x^3-3 x^2+3 x-2=0 \\
\Rightarrow & 2\left(x^3-1\right)-3 x(x-1)=0 \\
\Rightarrow & 2(x-1)\left(x^2+x+1\right)-3 x(x-1)=0 \\
\Rightarrow & (x-1)\left[2 x^2+2 x+2-3 x\right]=0 \\
\Rightarrow & (x-1)\left[2 x^2-x+2\right]=0 \\
x=1 \text { is real root. }
\end{array}
$$

So, sum of non-real complex root from the quadratic equation $2 x^2-x+2=0$ is $-\left(\frac{-1}{2}\right)=\frac{1}{2}$ Hence, option (a) is correct.