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(D) Given that $\overrightarrow{a}, \overrightarrow{b}$,and $\overrightarrow{c}$ are non-coplanar vectors,their scalar triple product is non-zero: $[\

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$-1$

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(D) Given that $\overrightarrow{a}, \overrightarrow{b}$,and $\overrightarrow{c}$ are non-coplanar vectors,their scalar triple product is non-zero: $[\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] = \left|\begin{array}{lll}1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2\end{array}ight| 
eq 0$. Let $\Delta = \left|\begin{array}{lll}1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2\end{array}ight|$. We are given the determinant equation: $\left|\begin{array}{lll}a & a^2 & 1+a^3 \ b & b^2 & 1+b^3 \ c & c^2 & 1+c^3\end{array}ight| = 0$. Using the property of determinants,we can split this into two determinants: $\left|\begin{array}{lll}a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1\end{array}ight| + \left|\begin{array}{lll}a & a^2 & a^3 \ b & b^2 & b^3 \ c & c^2 & c^3\end{array}ight| = 0$. In the second determinant,take $a, b, c$ common from rows $1, 2, 3$ respectively: $\left|\begin{array}{lll}a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1\end{array}ight| + abc \left|\begin{array}{lll}1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2\end{array}ight| = 0$. Note that $\left|\begin{array}{lll}a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1\end{array}ight| = \left|\begin{array}{lll}1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2\end{array}ight| = \Delta$ (by swapping columns twice). Thus,$\Delta + abc \Delta = 0 \Rightarrow \Delta(1 + abc) = 0$. Since $\Delta 
eq 0$,we must have $1 + abc = 0$,which implies $abc = -1$.

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