If a, b, c are distinct and left| begin{array | vedclass

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(B) Given the determinant equation: $\left| \begin{array}{ccc} a & a^2 & a^3 - 1 \ b & b^2 & b^3 - 1 \ c & c^2 & c^3 - 1 \end{array} ight| = 0$ We

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$abc = 1$

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(B) Given the determinant equation: $\left| \begin{array}{ccc} a & a^2 & a^3 - 1 \ b & b^2 & b^3 - 1 \ c & c^2 & c^3 - 1 \end{array} ight| = 0$ We can split the determinant into two: $\left| \begin{array}{ccc} a & a^2 & a^3 \ b & b^2 & b^3 \ c & c^2 & c^3 \end{array} ight| - \left| \begin{array}{ccc} a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1 \end{array} ight| = 0$ Taking $abc$ common from the first determinant and rearranging columns in the second: $abc \left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight| - \left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight| = 0$ Factoring out the common determinant: $(abc - 1) \left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight| = 0$ Since $a, b, c$ are distinct,the Vandermonde determinant $\left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight| = (a-b)(b-c)(c-a) 
eq 0$. Therefore,$abc - 1 = 0$,which implies $abc = 1$.

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

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