Question

If $a\ne b\ne c$ such that
$\left|\begin{array}{ccc}{a}^3-1& b^3-1& c^3-1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=0$ then,

• A. ab + bc + ca = 0
• B. a + b + c = 0
• C. abc = 1
• D. a + b + c =1

Answer 1

The correct option is C

abc = 1

$\Delta =\left|\begin{array}{ccc}{a}^3-1& b^3-1& c^3-1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=\left|\begin{array}{ccc}{a}^3& b^3& c^3\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|+\left|\begin{array}{ccc}-1& -1& -1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$
$\Delta =abc\left|\begin{array}{ccc}{a}^2& b^2& c^2\\ 1& 1& 1\\ a& b& c\end{array}\right|+\left|\begin{array}{ccc}-1& -1& -1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|\Delta =abc\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|-1\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|$
$\Delta =(abc-1)\left|\begin{array}{ccc}1& 1& 1\\ a& b& c\\ a^2& b^2& c^2\end{array}\right|=(abc-1)(a-b)(b-c)(c-a)Since,a\ne b\ne c,\Delta =0\Rightarrow (abc-1)=0\Rightarrow abc=1$