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$\overrightarrow{A}=1,a,a^2,\overrightarrow{B}=1,b,b^2,\overrightarrow{C}=1,c,c^2$are non-coplanar, then find the value of product $abc$

• A. -1
• B. 0
• C. 1
• D. None of these

Answer 1

The correct option is A -1

Let $D=\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|+\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& c^3\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|+abc\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|(1+abc)=0$ (i)
Since the vectors $(1,a,a^2),(1,b,b^2),(1,c,c^2)$ are non-coplanar, we have $\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|\ne 0$
Then (1) gives $abc=-1.$