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Q.
If $a, b, c$ are distinct and $\begin{vmatrix}a & a^2 & a^3-1 \\ b & b^2 & b^3-1 \\ c & c^2 & c^3-1\end{vmatrix}=0$ then $abc$ equals
Determinants
Solution:
Write the determinant as $=a b c \Delta_1-\Delta_2$
where $\Delta_1=\begin{vmatrix}1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2\end{vmatrix}=(a-b)(b-c)(c-a)$
and $\Delta_2=\begin{vmatrix}a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1\end{vmatrix}=\Delta_1$
$\therefore(a-b)(b-c)(c-a)(a b c-1)=0$
Since, $a, b, c$ are distinct, we get
$a b c-1=0 \text { or } a b c=1$