If vmatrixa&a^2&1+a^3&b^2&1+b^3&c^2&1+c^3vmatrix=0 and the vectors (1,a,a^2), (1,b,b^2) an…

Question

If $\begin{vmatrix}a&a^2&1+a^3\b&b^2&1+b^3\c&c^2&1+c^3\end{vmatrix}=0$ and the vectors $(1,a,a^2)$, $(1,b,b^2)$ and $(1,c,c^2)$ are non-coplanar, then the product $abc$ equals:

Accepted Answer

Splitting the determinant:

$$\begin{vmatrix}a&a^2&1\b&b^2&1\c&c^2&1\end{vmatrix}+\begin{vmatrix}a&a^2&a^3\b&b^2&b^3\c&c^2&c^3\end{vmatrix}=0 \implies (1+abc)\begin{vmatrix}1&a&a^2\1&b&b^2\1&c&c^2\end{vmatrix}=0$$

Since $(1,a,a^2),(1,b,b^2),(1,c,c^2)$ are non-coplanar, the remaining determinant is nonzero, so $1+abc=0 \implies abc=-1$

Solutions · 1