The vectors overline{p}=hat{i}+a hat{j}+a^2 h | vedclass

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

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(B) Given the determinant equation:$\left|\begin{array}{lll} a & a^2 & 1+a^3 \ b & b^2 & 1+b^3 \ c & c^2 & 1+c^3 \end{array}ight|=0$This can be split into two determinants:$\left|\begin{array}{lll} a & a^2 & 1 \ b & b^2 & 1 \ c & c^2 & 1 \end{array}ight| + \left|\begin{array}{lll} a & a^2 & a^3 \ b & b^2 & b^3 \ c & c^2 & c^3 \end{array}ight| = 0$Taking $abc$ common from the second determinant:$\left|\begin{array}{lll} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array}ight| + abc \left|\begin{array}{lll} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array}ight| = 0$$(1 + abc) \left|\begin{array}{lll} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array}ight| = 0$The determinant $\left|\begin{array}{lll} 1 & a & a^2 \ 1 & b & b^2 \ 1 & c & c^2 \end{array}ight|$ is the Vandermonde determinant,which equals $(a-b)(b-c)(c-a)$.Since the vectors $\overline{p}, \overline{q}, \overline{r}$ are non-coplanar,their scalar triple product is non-zero,which implies $a, b, c$ are distinct,so $(a-b)(b-c)(c-a) 
eq 0$.Therefore,$1 + abc = 0$,which gives $abc = -1$.

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