Suppose $a, b, c$ and $x$ are real numbers. Let $\mathrm{\Delta }=\left|\begin{array}{ccc}1+a& 1+ax& 1+a{x}^2\\ 1+b& 1+bx& 1+b{x}^2\\ 1+r& 1+rx& 1+r{x}^2\end{array}\right|$ .Then $∆$ is independent of

Write $\mathrm{\Delta }={\mathrm{\Delta }}_1+{\mathrm{\Delta }}_2,$ where

        ${\mathrm{\Delta }}_1=\left|\begin{array}{ccc}1& 1+ax& 1+a{x}^2\\ 1& 1+bx& 1+b{x}^2\\ 1& 1+cx& 1+c{x}^2\end{array}\right|$

and       ${\mathrm{\Delta }}_2=\left|\begin{array}{l}a\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}ax\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a{x}^2\\ b\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}bx\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}b{x}^2\\ c\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}cx\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}c{x}^2\end{array}\right|=0$

                 $\left[\because C_1\text{ and }{C}_2\text{ are propotional }\right]$

In ${\mathrm{\Delta }}_1,$use $C_2\to C_2-C_1,C_3\to C_3-C_1$ to obtain

${\mathrm{\Delta }}_1=\left|\begin{array}{ccc}1& ax& a{x}^2\\ 1& bx& b{x}^2\\ 1& cx& c{x}^2\end{array}\right|=0$

                     $\left[\because C_2\text{ and }{C}_3\text{ are propotional }\right]$

Thus $\mathrm{\Delta }=0$ and hence independence $a,b,c,x\text{. }$