Prove that the value of $(abc)-(ab+bc+ca)+3(a+b+c)$ is $0$ If the points $\big(\frac{a^3}{a-1}, \frac{a^2-3}{a-1}),(\frac{b^3}{b-1}, \frac{b^2-3}{b-1}) ,\big(\frac{c^3}{c-1}, \frac{c^2-3}{c-1}\big)$ are collinear for three distinct values of $a,b,c$ and $a,b,c\neq1$, then prove that the value of $(abc)-(ab+bc+ca)+3(a+b+c)$ is $0$.
What should be the smart approach to this question. If I try equating area of triangle formed by three points equal to $0$, then it gets too complicated. Please provide some insight.

HINT.- I do not  know if my answer could be “the smart approach to this question” as you want. Anyway here I give you the following.
Equalizing slopes
$$\frac{\frac{a^3-3}{a-1}-\frac{b^3-3}{b-1}}{\frac{a^3}{a-1}-\frac{b^3}{b-1}}=1-\frac{\frac{3}{b-1}-\frac{3}{a-1}}{\frac{a^3}{a-1}-\frac{b^3}{b-1}}=1-\frac{3}{ab(a+b)-(a^2+ab+b^2)}$$
So, by symmetry,  $$a^2b+ab^2-(a^2+ab+b^2)=a^2c+ac^2-(a^2+ac+c^2)=b^2c+bc^2-(b^2+bc+c^2)$$
It follows $$\begin{cases}\color{green}{(a^2b+ab^2)}+(a^2+ac+c^2)=a^2c+ac^2+\color{green}{(a^2+ab+b^2)}\\\color{green}{(a^2b+ab^2)}+(b^2+bc+c^2)=b^2c+bc^2+\color{green}{(a^2+ab+b^2)}\end {cases}\qquad(*)$$
Hence, by subtracting the equalities in (*),
$$(a^2+ac+c^2)- (b^2+bc+c^2)= (a^2c+ac^2)-(bc^2+bc^2)\Rightarrow(c-1)(a+b+c)=0$$ 
Thus, because of $c\ne 1$, $$\color {red}{a+b+c=0}$$ 
Consequently you have to prove the easier relation
$$abc-(ab+ac+bc)=0$$
There are several ways to solve now. Try to finish the question. If you cannot  (I believe you can), I will return.
