Question

If $\left|\begin{array}{c}a{a}^2{a}^3-1\\ b{b}^2{b}^3-1\\ c{c}^2{c}^3-1\end{array}\right|=0$ in which $a,b,c$ are different, show that $abc=1$.

Answer 1

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

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

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

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

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

$\therefore abc=1$