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 vectors $(1, a, a^2) (1, b, b^2)$ and $(1, c, c^2)$ are non-coplanar, then the product abc equals

• A. $2$
• B. $-1$
• C. $1$
• D. $0$

Answer 1

Answer: (B)

$-1$

Answer 2

$\begin{vmatrix}a&a^{2}&1\\\ b&b^{2}&1\\\ c&c^{2}&1\end{vmatrix}+\begin{vmatrix}1&a&a^{2}\\\ 1&b&b^{2}\\\ 1&c&c^{2}\end{vmatrix}=0$ $\left(1+abc\right)\begin{vmatrix}a&a^{2}&1\\\ b&b^{2}&1\\\ c&c^{2}&1\end{vmatrix}=0$ $\Rightarrow abc=-1.$ Hence, (B) is the correct answer