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
$\overline{A}=(1,a,a^2)$
$\overline{B}=(1,b,b^2)$
$\overline{C}=(1,c,c^2)$
are non coplanar then the value of $abc$ equals

• A. $2$
• B. $0$
• C. $-1$
• D. $-2$

Answer 1

The correct option is C $-1$

$\left|\begin{array}{ccc}a& a^2& 1+a^3\\ b& b^2& 1+b^3\\ c& c^2& 1+c^3\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right|+\left|\begin{array}{ccc}a& a^2& a^3\\ b& b^2& b^3\\ c& c^2& c^3\end{array}\right|=0$
$\Rightarrow (1+abc)\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=0$
$\Rightarrow 1+abc=0as\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|\ne 0,(\because \overline{A},\overline{B},\overline{C}arenoncoplanar)$
$\therefore abc=-1$