Question

If $a,b,c$ are unequal and different from one such that the points $(\frac{a^3}{a-1},\frac{a^2-3}{a-1}),(\frac{b^3}{b-1},\frac{b^2-3}{b-1})$ and $(\frac{c^3}{c-1},\frac{c^3-3}{c-1})$ are collinear, then

• A. $bc+ca+ab+abc=0$
• B. $a+b+c=abc$
• C. $bc+ca+ab=abc$
• D. $bc+ca+ab-abc=3(a+b+c)$

Answer 1

The correct option is D $bc+ca+ab-abc=3(a+b+c)$

Given points are $(\frac{a^3}{a-1},\frac{a^2-3}{a-1}),(\frac{b^3}{b-1},\frac{b^2-3}{b-1}),(\frac{c^3}{c-1},\frac{c^2-3}{c-1})$

They are collinear if area of triangle formed by these point is equal to $0.$

$\left|\begin{array}{ccc}\frac{a^3}{a-1}& \frac{a^2-3}{a-1}& 1\\ \frac{b^3}{b-1}& \frac{b^2-3}{b-1}& 1\\ \frac{c^3}{c-1}& \frac{c^2-3}{c-1}& 1\end{array}\right|=0|$

$⟹(a-1)(b-1)(c-1)\left|\begin{array}{ccc}{a}^3& a^2-3& a-1\\ b^3& b^2-3& b-1\\ c^3& c^2-3& c-1\end{array}\right|=0$

solving the above determinant we get

$abc-ab-ac+3a+b^{2}c-b^2-bc+3b+b{c}^2-bc-c^2+3c-b^{2}c+b^2-b{c}^2+bc+c^2=0$

$⟹abc-(ab+bc+ca)+3(a+b+c)=0$

$⟹bc+ca+ab-abc=3(a+b+c)$