If a, b, c are all different and the equations
$a x + a^{2} y + (a^{3} + 1) = 0$
$b x + b^{2} y + (b^{3} + 1) = 0$
$c x + c^{2} y + (c^{3} + 1) = 0$
are consistent then prove that $a b c + 1 = 0$

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Solution

For consistency $D = 0$

$∴ ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & a^{2} & a^{3} + 1 b & b^{2} & b^{3} + 1 c & c^{2} & c^{3} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

Hence we have $1 + a b c = 0 .$

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