If the points $(\frac{a^{3}}{a - 1}, \frac{a^{2} - 3}{a - 1}), (\frac{b^{3}}{b - 1}, \frac{b^{2} - 3}{b - 1}), (\frac{c^{3}}{c - 1}, \frac{c^{2} - 3}{c - 1})$ are collinear for three distinct values of $a, b, c$ , then show that $a b c - (b c + c a + a b) + 3 (a + b + c) = 0.$

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Solution

It is given that three points $(\frac{a^{3}}{a - 1}, \frac{a^{3} - 3}{a - 1}), (\frac{b^{3}}{b - 1}, \frac{b^{3} - 3}{b - 1}), (\frac{c^{3}}{c - 1}, \frac{c^{3} - 3}{c - 1})$ are collinear.
$a, b, c$ are distinct and $a \neq 1, b \neq 1, c \neq 1$
$(x_{1} y_{1}), (x_{2} y_{2})$ & $(x_{3} y_{3})$ are collinear if
$Δ = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$
Using this property,
$\frac{1}{2} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} \frac{a^{3}}{a^{2} - 1} & \frac{(a^{3} - 3)}{(a - 1)} & 1 \frac{b^{3}}{b^{2} - 1} & \frac{(b^{2} - 3)}{(b - 1)} & 1 \frac{c^{3}}{c^{2} - 1} & \frac{(c^{2} - 3)}{(c - 1)} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 0 \Rightarrow (a - 1) (b - 1) (c - 1) ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{3} & a^{2} - 3 & a - 1 b^{3} & b^{2} - 3 & b - 1 c^{3} & c^{2} - 3 & c - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$
$\Rightarrow ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{3} & a^{2} - 3 & a - 1 b^{3} & b^{2} - 3 & b - 1 c^{3} & c^{2} - 3 & c - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 (∵ a \neq b \neq c \neq 1, (a - 1) \neq (b - 1) \neq (c - 1) \neq 0) R_{1} \leftrightarrow R_{1} - R_{2}, R_{2} \leftrightarrow R_{2} - R_{3} \Rightarrow ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{3} - b^{3} & a^{2} - b^{2} & a - b b^{3} - c^{3} & b^{2} - c^{2} & b - c c^{3} & c^{3} - 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 \Rightarrow (a - b) (b - c) ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + b^{2} + a b & a + b & 1 b^{2} + c^{2} + b c & b + c & 1 c^{3} & c^{2} - 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 R_{1} \leftarrow R_{1} - R_{2}$
$\Rightarrow ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} - c^{2} + b (a - c) & a - c & 0 b^{2} + c^{2} + b c & b + c & 1 c^{3} & c^{2} - 3 & c - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 ((a - b) = (b - c) \neq 0, a, b, c$ are distinct $)$
$\Rightarrow (a - c) ∣ ∣ ∣ ∣ ∣ \begin{matrix} a + c + b & 1 & 0 b^{2} + c^{2} + b c & b + c & 1 c^{3} & c^{2} - 3 & c - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 \Rightarrow ∣ ∣ ∣ ∣ ∣ \begin{matrix} a + c + b & 1 & 0 b^{2} + c^{2} + b c & b + c & 1 c^{3} & c^{2} - 3 & c - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$
Expanding along the ${I I I}^{r d}$ column we have,
$\Rightarrow (c - 1) {(a + b) (b + c) - b^{2}} - {(a + b) (c^{2} - 3) - 3 c} = 0 \Rightarrow (c - 1) (a b + b^{2} + a c + b c - b^{2}) - (a c^{2} - 3 a + b c^{2} - 3 b - 3 c) = 0 \Rightarrow (c - 1) (a b + b c + c a) - [(a + b) c^{2} - 3 (a + b + c)] = 0 \Rightarrow a b c + {b c}^{2} + {a c}^{2} - a b - b c - c a - {a c}^{2} - {b c}^{2} + 3 (a + b + c) = 0 \Rightarrow a b c - (a b + b c + c a) + 3 (a + b + c) = 0$

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