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 $A = (1, a, a^{2}), B= (1, b, b^{2}), C=(1, c, c^{2})$ are non-coplanar, then the value of abc is equal to

• A. $-2$
• B. $0$
• C. $1$
• D. $-1$

Answer 1

Answer: (D)

Given, $\begin{vmatrix}a&a^{2}&1+a^{3}\\ b&b^{2}&1+b^{3}\\ c&c^{2}&1+c^{3}\end{vmatrix}=0 $
$\Rightarrow \begin{vmatrix}a&a^{2}&1\\ b&b^{2}&1\\ c&c^{2}&1\end{vmatrix}+abc \begin{vmatrix}1&a&a^{2}\\ 1&b&b^{2}\\ 1&c&c^{2}\end{vmatrix}=0 $
$\Rightarrow \left(abc+1\right)\begin{vmatrix}1&a&a^{2}\\ 1&b&b^{2}\\ 1&c&c^{2}\end{vmatrix}=0$
$\Rightarrow \left(abc+1\right)\begin{vmatrix}1&a&a^{2}\\ 1&b&b^{2}\\ 1&c&c^{2}\end{vmatrix}=0 $
Since given three vectors are non coplanar
$\therefore abc =-1$