If the points$({\mathrm{a}}^3/(\mathrm{a}-1),({\mathrm{a}}^2-3)/(\mathrm{a}-1\left)\right),({\mathrm{b}}^3/(\mathrm{b}-1)$ and $({\mathrm{c}}^3/(\mathrm{c}-1),({\mathrm{c}}^2-3)/(\mathrm{c}-1\left)\right)$ where a, b, c are different from 1, lie on the line 

lx + my + n=0, then

Since the given points lie on the line lx + my + n= 0, a, b, c are the roots of the equation

$\begin{array}{rl} & \mathrm{l}\left(\frac{{\mathrm{t}}^3}{\mathrm{t}-1}\right)+\mathrm{m}\left(\frac{{\mathrm{t}}^2-3}{\mathrm{t}-1}\right)+\mathrm{n}=0\\ \mathrm{or} & {\mathrm{lt}}^3+{\mathrm{mt}}^2+\mathrm{nt}-(3\mathrm{m}+\mathrm{n})=0----\left(1\right)\end{array}$

Hence,

$\begin{array}{rl} & \mathrm{a}+\mathrm{b}+\mathrm{c}=-\frac{\mathrm{m}}{\mathrm{l}}\\ & \mathrm{ab}+\mathrm{bc}+\mathrm{ca}=\frac{\mathrm{n}}{\mathrm{l}}----\left(2\right)\end{array}$

$\mathrm{abc}=\frac{3\mathrm{m}+\mathrm{n}}{\mathrm{l}}$

So, from (1), (2), and (3), we get

$\mathrm{abc}-(\mathrm{bc}+\mathrm{ca}+\mathrm{ab})+3(\mathrm{a}+\mathrm{b}+\mathrm{c})=0$