Question

Prove the following :
$\left|\begin{array}{cccc}{x}^3& {3x}^2& 3x& 1\\ x^2& x^2+2x& 2x+1& 1\\ x& 2x+1& x+2& 1\\ 1& 3& 3& 1\end{array}\right|={(x-1)}^6.$

Answer 1

Keeping in view that $x^3-3x^2+3x-1=(x-1{)}^3$,
we apply $C_1-C_2+C_3-C_4$.
$\therefore \Delta =\left|\begin{array}{cccc}{(x-1)}^3& {3x}^2& 3x& 1\\ 0& x^2+2x& 2x+1& 1\\ 0& 2x+1& x+2& 1\\ 0& 3& 3& 1\end{array}\right|$
$={(x-1)}^3\left|\begin{array}{ccc}{x}^2+2x& 2x+1& 1\\ 2x+1& x+2& 1\\ 3& 3& 1\end{array}\right|$
Now make two zeros by $R_1-R_2$ and $R_2-R_3$
$\Delta ={(x-1)}^3\left|\begin{array}{ccc}{x}^2-1& x-1& 0\\ 2(x-1)& x-1& 1\\ 3& 3& 1\end{array}\right|$
$=(x-1{)}^3(x-1{)}^2\left|\begin{array}{cc}x+1& 1\\ 2& 1\end{array}\right|$
$=(x-1{)}^5(x+1-2)=(x-1{)}^6$.