Question

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