If a,b,c are unequal and different from 1 such that the points $\left(\frac{{\mathrm{a}}^3}{\mathrm{a}-1},\frac{{\mathrm{a}}^3-3}{\mathrm{a}-1}\right)$,$\left(\frac{{\mathrm{b}}^3}{\mathrm{b}-1},\frac{{\mathrm{b}}^3-3}{\mathrm{b}-1}\right),\left(\frac{{\mathrm{c}}^3}{\mathrm{c}-1},\frac{{\mathrm{c}}^3-3}{\mathrm{c}-1}\right)$ are collinear then

Suppose the given points lies on the line $\mathrm{lx}+\mathrm{my}+\mathrm{n}=0$, Then a,b,c are the roots of the equation  ${\mathrm{lt}}^3+\mathrm{m}\left({\mathrm{t}}^2-3\right)+\mathrm{n}\left(\mathrm{t}-1\right)=0$ $\Rightarrow$ ${\mathrm{lt}}^3+{\mathrm{mt}}^2+\mathrm{nt}-\left(3\mathrm{m}+\mathrm{n}\right)=0$
The relations between the roots and coefficients are as below
Sum of the roots 
 $\mathrm{a}+\mathrm{b}+\mathrm{c}=-\frac{\mathrm{m}}{\mathrm{l}}$       …… (1) 
Sum of the product of the roots taken two at a time 
  $\mathrm{bc}+\mathrm{ca}+\mathrm{ab}=\frac{\mathrm{n}}{\mathrm{l}}$       …… (2)
The product of the roots is 
 $\mathrm{abc}=\frac{\left(3\mathrm{m}+\mathrm{n}\right)}{\mathrm{l}}$       …… (3)   
Eliminating  $\mathrm{l},\mathrm{m},\mathrm{n}$  we get
 $\begin{array}{l}\mathrm{abc}=\frac{3\mathrm{m}}{\mathrm{l}}+\frac{\mathrm{n}}{\mathrm{l}}\\ =-3\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)+\mathrm{bc}+\mathrm{ca}+\mathrm{ab}\end{array}$
Therefore, the condition is   $\boxed{\mathrm{bc}+\mathrm{ca}+\mathrm{ab}-\mathrm{abc}=3\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}$