Given $n_1, n_2, n_3$ are roots of the equation
$E_1: x^3+x^2+l x+n=0$ $\ldots(\mathrm{i})$
$\therefore \text { Sum of roots } x_1+x_2+x_3=-1$ $\ldots(\mathrm{ii})$
Now, $E_2: x^3+a x^2+b x+c=0$ is reciprocal equation of class one $\Rightarrow c=1$ and $a=b$
$\therefore \quad E_2: x^3+a x^2+a x+1=0$ $\ldots(\mathrm{iii})$
Given, Eq. (iii) have roots
$\frac{x_1-1}{2}, \frac{x_2-1}{2}, \frac{x_3-1}{2}$
$\Rightarrow \quad \frac{x_1-1}{2}+\frac{x_2-1}{2}+\frac{x_3-1}{2}=-a$
$\Rightarrow \quad \frac{x_1+x_2+x_3}{2}-\frac{3}{2}=-a$
From Eq. (ii),
$x_1+x_2+x_3=-1$
$\therefore \quad \frac{-1}{2}-\frac{3}{2}=-a \Rightarrow a=2$
$\therefore$ Eq. (iii), $x^3+2 x^2+2 x+1=0$
$(x+1)\left(x^2+x+1\right)=0 \quad \therefore \quad x=-1, x^2+x+1=0$
$\Rightarrow \quad x=\frac{-1 \pm \sqrt{3} i}{2}$
$\therefore$ Given, $\frac{x_1-1}{2}=-1 \Rightarrow x_1=-1$
$\frac{x_2-1}{2}=\frac{-1+\sqrt{3} i}{2} \Rightarrow x_2=\sqrt{3} i$
$\frac{x_3-1}{2}=\frac{-1-\sqrt{3} i}{2} \Rightarrow x_3=-\sqrt{3} i$
$\therefore$ Roots of these two equation excluding common roots are
$\sqrt{3} i,-\sqrt{3} i, \frac{-1+\sqrt{3} i}{2}, \frac{-1-\sqrt{3} i}{2}$