Question:

Step 1: Decompose the given determinant by separating the third column. $$ \begin{vmatrix} a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3 \end{vmatrix} = \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} + \begin{vmatrix} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{vmatrix} = 0 $$ Step 2: Factor the second determinant by extracting common factors from each row. $$ \begin{vmatrix} a & a^2 & a^3 \\ b & b^2 & b^3 \\ c & c^2 & c^3 \end{vmatrix} = abc\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} $$ Step 3: Relate the first determinant to the Vandermonde-type determinant. Let $D = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix}$. The first determinant can be transformed into $D$ by two column swaps: $$ \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} = - \begin{vmatrix} 1 & a^2 & a \\ 1 & b^2 & b \\ 1 & c^2 & c \end{vmatrix} \quad (\text{swap } C_1 \leftrightarrow C_3) $$ $$ - \begin{vmatrix} 1 & a^2 & a \\ 1 & b^2 & b \\ 1 & c^2 & c \end{vmatrix} = - \left( - \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \right) \quad (\text{swap } C_2 \leftrightarrow C_3) $$ Thus, $$ \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = D $$ Step 4: Substitute the expressions back into the decomposed equation. $$ D + abc \cdot D = 0 $$ Factor out $D$: $$ D(1 + abc) = 0 $$ Step 5: Apply the condition that the vectors are non-coplanar. The vectors $(1, a, a^2)$, $(1, b, b^2)$, and $(1, c, c^2)$ are non-coplanar. This implies that the determinant formed by these vectors is non-zero. $$ D = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \neq 0 $$ Step 6: Solve for $abc$. Since $D \neq 0$, for the product $D(1 + abc)$ to be zero, we must have: $$ 1 + abc = 0 $$ $$ abc = -1 $$