# show that determinant- | x2 2x 1 | is a perfect square 1 x2 2x 2x 1 x2

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Please find below the solution to the asked query:
$∆=\left|\begin{array}{ccc}{\mathrm{x}}^{2}& 2\mathrm{x}& 1\\ 1& {\mathrm{x}}^{2}& 2\mathrm{x}\\ 2\mathrm{x}& 1& {\mathrm{x}}^{2}\end{array}\right|\phantom{\rule{0ex}{0ex}}{\mathrm{C}}_{1}\to {\mathrm{C}}_{1}+{\mathrm{C}}_{2}+{\mathrm{C}}_{3}\phantom{\rule{0ex}{0ex}}⇒∆=\left|\begin{array}{ccc}{\mathrm{x}}^{2}+2\mathrm{x}+1& 2\mathrm{x}& 1\\ {\mathrm{x}}^{2}+2\mathrm{x}+1& {\mathrm{x}}^{2}& 2\mathrm{x}\\ {\mathrm{x}}^{2}+2\mathrm{x}+1& 1& {\mathrm{x}}^{2}\end{array}\right|\phantom{\rule{0ex}{0ex}}=\left|\begin{array}{ccc}{\left(\mathrm{x}+1\right)}^{2}& 2\mathrm{x}& 1\\ {\left(\mathrm{x}+1\right)}^{2}& {\mathrm{x}}^{2}& 2\mathrm{x}\\ {\left(\mathrm{x}+1\right)}^{2}& 1& {\mathrm{x}}^{2}\end{array}\right|\phantom{\rule{0ex}{0ex}}\mathrm{Take}{\left(\mathrm{x}+1\right)}^{2}\mathrm{common}\mathrm{from}{\mathrm{C}}_{1}\phantom{\rule{0ex}{0ex}}∆={\left(\mathrm{x}+1\right)}^{2}\left|\begin{array}{ccc}1& 2\mathrm{x}& 1\\ 1& {\mathrm{x}}^{2}& 2\mathrm{x}\\ 1& 1& {\mathrm{x}}^{2}\end{array}\right|\phantom{\rule{0ex}{0ex}}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{3}\phantom{\rule{0ex}{0ex}}{\mathrm{R}}_{3}\to {\mathrm{R}}_{3}-{\mathrm{R}}_{1}\phantom{\rule{0ex}{0ex}}∆={\left(\mathrm{x}+1\right)}^{2}\left|\begin{array}{ccc}1& 2\mathrm{x}& 1\\ 1-1& {\mathrm{x}}^{2}-1& 2\mathrm{x}-{\mathrm{x}}^{2}\\ 1-1& 1-2\mathrm{x}& {\mathrm{x}}^{2}-1\end{array}\right|\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}={\left(\mathrm{x}+1\right)}^{2}\left|\begin{array}{ccc}1& 2\mathrm{x}& 1\\ 0& {\mathrm{x}}^{2}-1& 2\mathrm{x}-{\mathrm{x}}^{2}\\ 0& 1-2\mathrm{x}& {\mathrm{x}}^{2}-1\end{array}\right|\phantom{\rule{0ex}{0ex}}\mathrm{Expand}\mathrm{along}{\mathrm{C}}_{1}\phantom{\rule{0ex}{0ex}}∆={\left(\mathrm{x}+1\right)}^{2}\left[1{\left({\mathrm{x}}^{2}-1\right)}^{2}-\left(2\mathrm{x}-{\mathrm{x}}^{2}\right)\left(1-2\mathrm{x}\right)\right]\phantom{\rule{0ex}{0ex}}={\left(\mathrm{x}+1\right)}^{2}\left[{\mathrm{x}}^{4}+1-2{\mathrm{x}}^{2}-\left(2\mathrm{x}-4{\mathrm{x}}^{2}-{\mathrm{x}}^{2}+2{\mathrm{x}}^{3}\right)\right]\phantom{\rule{0ex}{0ex}}={\left(\mathrm{x}+1\right)}^{2}\left[{\mathrm{x}}^{4}+1-2{\mathrm{x}}^{2}-2\mathrm{x}+4{\mathrm{x}}^{2}+{\mathrm{x}}^{2}-2{\mathrm{x}}^{3}\right]\phantom{\rule{0ex}{0ex}}={\left(\mathrm{x}+1\right)}^{2}\left({\mathrm{x}}^{4}-2{\mathrm{x}}^{3}+3{\mathrm{x}}^{2}-2\mathrm{x}+1\right)\phantom{\rule{0ex}{0ex}}={\left(\mathrm{x}+1\right)}^{2}\left({\mathrm{x}}^{4}+1+{\mathrm{x}}^{2}-2{\mathrm{x}}^{3}+2{\mathrm{x}}^{2}-2\mathrm{x}\right)\phantom{\rule{0ex}{0ex}}={\left(\mathrm{x}+1\right)}^{2}\left({\mathrm{x}}^{4}+1+{\mathrm{x}}^{2}+2{\mathrm{x}}^{2}-2\mathrm{x}-2{\mathrm{x}}^{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Using}{\left(\mathrm{a}-\mathrm{b}-\mathrm{c}\right)}^{2}={\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}+2\mathrm{ab}-2\mathrm{bc}-2\mathrm{ca},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}△={\left(\mathrm{x}+1\right)}^{2}{\left({\mathrm{x}}^{2}+1-\mathrm{x}\right)}^{2}\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{perfect}\mathrm{square}.\left(\mathrm{Hence}\mathrm{Proved}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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