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(A) Given that the vectors $(1, a, a^2)$,$(1, b, b^2)$,and $(1, c, c^2)$ are non-coplanar,the determinant of the matrix formed by these vectors is non

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$-1$

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(A) Given that the vectors $(1, a, a^2)$,$(1, b, b^2)$,and $(1, c, c^2)$ are non-coplanar,the determinant of the matrix formed by these vectors is non-zero:$\Delta = \left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight| 
eq 0$.We are given the equation:$\left| \begin{array}{ccc} a & a^2 & 1 + a^3 \ b & b^2 & 1 + b^3 \ c & c^2 & 1 + c^3 \end{array} ight| = 0$.Using the property of determinants,we can split the third column:$\left| \begin{array}{ccc} a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1 \end{array} ight| + \left| \begin{array}{ccc} a & a^2 & a^3 \ b & b^2 & b^3 \ c & c^2 & c^3 \end{array} ight| = 0$.In the second determinant,factor out $a, b, c$ from the rows:$\left| \begin{array}{ccc} a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1 \end{array} ight| + abc \left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight| = 0$.Note that $\left| \begin{array}{ccc} a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1 \end{array} ight| = \left| \begin{array}{ccc} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array} ight| = \Delta$ (after two column swaps).Thus,$\Delta + abc \Delta = 0 \Rightarrow \Delta(1 + abc) = 0$.Since $\Delta 
eq 0$,we have $1 + abc = 0$,which implies $abc = -1$.

If $\left| \begin{array}{ccc} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \end{array} \right| = 0$ and the vectors $\vec{a} = (1, a, a^2)$,$\vec{b} = (1, b, b^2)$,and $\vec{c} = (1, c, c^2)$ are non-coplanar,then $abc$ is equal to

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