Prove that the lines
$(b + c) x - b c y = a (b^{2} + c^{2} + b c)$ ,
$(c + a) x - c a y = b (c^{2} + a^{2} + c a)$ ,
$(a + b) x - a b y = c (a^{2} + b^{2} + a b)$ ,
are concurrent at the point $(Σ a b, Σ a)$ where
$a, b, c$ are distinct real numbers.

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Solution

Multiplying the $1 s t$ equation by $a^{2} (b - c)$ , $2 n d$ by $b^{2} (c - a)$ and $3 r d$ $c^{2} (a - b)$ , their equations become
$a^{2} (b^{2} - c^{2}) x - a^{2} b c (b - c) y - a^{3} (b^{3} - c^{3}) = 0$
And so on. They will be concurrent if $△ = 0$ as in part $(a)$ .
Point of intersection:
Subtract $1 s t$ two and cancel $(b - a)$ .
$∴ x - c y = a b - c^{2}$
Or $x = c y + (a b - c^{2})$
Putting for $x$ in any and simplifying, we get $y = a + b + c$ and hence $x = a b + b c + c a$ , by $(1)$ .

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