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

Answer 1

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

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

Taking out common $a,,b,c$ from $R_1,R_2,R_3$ respectively from second determent.

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

Replacing $C_1$ and $C_2$ and then $C_2$ and $C_3$

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

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

$abc+1=0$

$abc=-1$ proved.