If a, b, c are distinct and aa2a3−1bb2b3−1cc2c3−1=0 ,then abc equals

If $a, b, c$ are distinct and $|\begin{array}{ccc} a & a^{2} & a^{3} - 1 \\ b & b^{2} & b^{3} - 1 \\ c & c^{2} & c^{3} - 1 \end{array}| = 0,$ then $a b c$ equals

Write the determinant as $= a b c Δ_{1} - Δ_{2}$

where $Δ_{1} = |\begin{array}{ccc} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{array}| = (a - b) (b - c) (c - a)$

and $Δ_{2} = |\begin{array}{ccc} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{array}| = Δ_{1}$

$∴ (a - b) (b - c) (c - a) (a b c - 1) = 0$

Since, $a, b, c$ are distinct, we get

$a b c - 1 = 0 or a b c = 1$

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