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.
 A: 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.
A: Let:
$$F=abc-(ab+bc+ca)+3(a+b+c) \ \ (0)$$
Let us consider the determinant              
$$D=\begin{vmatrix}\frac{a^3}{a-1}&\frac{a^2-3}{a-1}&1\\\frac{b^3}{b-1}&\frac{b^2-3}{‌​b-1}&1\\ \frac{c^3}{c-1}& \frac{c^2-3}{c-1}&1\end{vmatrix}=k\begin{vmatrix}a^3&a^2-3&a-1\\b^3&b^2-3&b-1\\ c^3&c^2-3&c-1\end{vmatrix} \ \ \ (1)$$
                     ‌​
with $k=\dfrac{1}{(a-1)(b-1)(c-1)}$.
(a method you have mentionned), equal to twice the area of the corresponding triangle. It is zero iff the corresponding points are collinear (mathworld.wolfram.com/Collinear.html). 
One can factorize $D$ (through a Computer Algebra system or by hand computation, by factoring  expressions $(a-c)$, $(b-c)$ and $(a-b)$, or by using the simpler method described at the bottom of this text), under the form:
$$D=F \times \frac{(a-b)(a-c)(b-c)}{(a-1)(b-1)(c-1)} \ \ (2)$$
with $F$ as defined by (0). So: $$D=0 \ \ \ \Leftrightarrow \ \ \ F=0$$
Edit : An important remark for problems of a similar kind 
I was puzzled by the fact to find back in (2) the classical factorization of the Vandermonde determinant: 
$$V=V(a,b,c)=\begin{vmatrix}a^2&a&1\\b^2&b&1\\ c^2&c&1\end{vmatrix}= (a-b)(a-c)(b-c) \ \ (3).$$ 
After some googling, I realized that the last determinant in formula (1) is a combination of "Generalized Vandermonde Determinants" (caution: the term can be employed with a different meaning), all having $V$ in factor.
It is a consequence of Theorem 1 in E.R. Heineman (1929): "Generalized Vandermonde Determinants" https://projecteuclid.org/download/pdf_1/euclid.bams/1183496754 generalized Vandermonde determinant heineman . 
It is maybe interesting to postpone the reading of this theorem now ; it is better to understand the key idea: the elementary symmetric polynomials in $n$ variables can be expressed, as generalized Vandermonde determinants divided by $V$. Here, for $n=3$:
$$\begin{vmatrix}a^3&a&1\\b^3&b&1\\ c^3&c&1\end{vmatrix} = V\times(a+b+c), \ \
\begin{vmatrix}a^3&a^2&1\\b^3&b^2&1\\ c^3&c^2&1\end{vmatrix} = V\times(ab+bc+ca), \ \ 
\begin{vmatrix}a^3&a^2&a\\b^3&b^2&b\\ c^3&c^2&c\end{vmatrix} = V\times(abc)
$$
Where $V=V(a,b,c)$ has been defined in (3).
Using these formulas, the development of the last determinant in (1) becomes very straightforward.
Remark: the rule behind the choices of these determinants is as follows: in the (left) augmented matrix:
$$\begin{bmatrix}a^2&a&1\\b^2&b&1\\ c^2&c&1\end{bmatrix} \ \ \rightarrow \ \ \begin{bmatrix}a^3&a^2&a&1\\b^3&b^2&b&1\\ c^3&c^2&c&1\end{bmatrix}
$$
delete, one at a time, one of the three last columns and take the determinant. This process is analogous in any dimension.
Bibliography: See also the more recent paper of Thorsten Werther: "Generalized Vandermonde Determinants over the Chebyshev Basis", 1993, International Computer Science Institute: ftp://ftp.icsi.berkeley.edu/pub/techreports/1993/tr-93-024.pdf 
