Question

If $a,b,c$ are all different and the points $(\frac{r^3}{r-1},\frac{r^2-3}{r-1})$ where $r=a,b,c$ are collinear, then show $3(a+b+c)=ab+bc+ca-abc.$

Answer 1

Consider a line $lx+my+n=0$

Point $(\frac{r^3}{r-1},\frac{r^2-3}{r-1})$ lies on the above line

$\therefore l\left(\frac{r^3}{r-1}\right)+m\left(\frac{r^2-3}{r-1}\right)+n=0$

$l{r}^3+m{r}^2+nr-(3m+n)=0$

And $a,b,c$ are the roots of the equation

$a+b+c=\frac{-m}{l},ab+bc+ca=\frac{n}{l},abc=\frac{3m+n}{l}$

Now taking LHS

$3(a+b+c)=\frac{-3m}{l}$

RHS

$ab+bc+ca-abc=\frac{n}{l}-\left(\frac{3m+n}{l}\right)=-\frac{3m}{l}$