If the vectors vec{alpha}=hat{i}+a hat{j}+a^{ | vedclass

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

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$-1$

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(C) Given the determinant equation: $\left|\begin{array}{lll}a & a^{2} & 1+a^{3} \ b & b^{2} & 1+b^{3} \ c & c^{2} & 1+c^{3}\end{array}ight|=0$ We can split the third column: $\left|\begin{array}{lll}a & a^{2} & 1 \ b & b^{2} & 1 \ c & c^{2} & 1\end{array}ight| + \left|\begin{array}{lll}a & a^{2} & a^{3} \ b & b^{2} & b^{3} \ c & c^{2} & c^{3}\end{array}ight| = 0$ Taking $abc$ common from the second determinant: $\left|\begin{array}{lll}a & a^{2} & 1 \ b & b^{2} & 1 \ c & c^{2} & 1\end{array}ight| + abc \left|\begin{array}{lll}a & a^{2} & 1 \ b & b^{2} & 1 \ c & c^{2} & 1\end{array}ight| = 0$ $(1+abc) \left|\begin{array}{lll}a & a^{2} & 1 \ b & b^{2} & 1 \ c & c^{2} & 1\end{array}ight| = 0$ Since $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are non-coplanar, the scalar triple product $[\vec{\alpha} \vec{\beta} \vec{\gamma}] 
eq 0$. The determinant $\left|\begin{array}{lll}1 & a & a^{2} \ 1 & b & b^{2} \ 1 & c & c^{2}\end{array}ight|$ is related to the scalar triple product. Since the rows/columns are swapped, the value is non-zero. Therefore, $1+abc = 0$, which implies $abc = -1$.

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