Question

If $a,b,c$ are unequal and different from $1$ 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 $(\frac{c^3}{c-1},\frac{c^2-3}{c-1})$ 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 point $(\frac{a^3}{a-1},\frac{a^2-3}{a-1}),(\frac{b^3}{b-1},\frac{b^2-3}{b-1})$ & $(\frac{c^3}{c-1},\frac{c^2-3}{c-1})$

They are collinear if area of $\Delta$le$=0$

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

$[\because a,b,c\ne 1]$

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

$R_1\to R_2;R_2-R_3$

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

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

$\Rightarrow$ Dividing by $(a-b)(b-c)$ on both sides & then dividing by $(a-c)$

$\Rightarrow [a-c]=\left|\begin{array}{ccc}a+b+c& 1& 0\\ b^2+bc+c^2& b+c& 1\\ c^3& c^2-3& c-1\end{array}\right|=0$

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

$\Rightarrow (a+b+c)\left|\begin{array}{cc}b+c& 1\\ c^2-3& c-1\end{array}\right|-1\left|\begin{array}{cc}{b}^2+bc+c^2& 1\\ c^3& c-1\end{array}\right|=0$

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

$\Rightarrow abc-(ab+bc+ca)+3(a+b+c)=0$

$\Rightarrow bc+ca+ab-abc=3(a+b+c)$