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 find the value of the product $abc$.

• A. 0
• B. 1
• C. -1
• D. None of the above

Answer 1

The correct option is C -1

Since $(1,a,a^2),(1,b,b^2),(1,c,c^2)$ are non coplanar
$\Delta =\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|\ne 0$
Given
$\left|\begin{array}{ccc}a& a^2& 1+a^2\\ b& b^2& 1+b^2\\ c& c^2& 1+c^2\end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|+\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& b^3\end{array}\right|=0$

$\Rightarrow (1+abc)\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=0$

$\Rightarrow (1+abc)\Delta =0$

As $\Delta \ne 0$, we get

$1+abc=0$

$\Rightarrow abc=-1$