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The value of the determinant $\left|\begin{array}{lll}\mathrm{m} & \mathrm{n} & \mathrm{p} \\ \mathrm{p} & \mathrm{m} & \mathrm{n} \\ \mathrm{n} & \mathrm{p} & \mathrm{m}\end{array}\right|$
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has linear factor
Consider $\left|\begin{array}{ccc}\mathrm{m} & \mathrm{n} & \mathrm{p} \\ \mathrm{p} & \mathrm{m} & \mathrm{n} \\ \mathrm{n} & \mathrm{p} & \mathrm{m}\end{array}\right|$
$\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$
$=\left|\begin{array}{ccc}\mathrm{m}+\mathrm{n}+\mathrm{p} & \mathrm{n} & \mathrm{p} \\ \mathrm{p}+\mathrm{m}+\mathrm{n} & \mathrm{m} & \mathrm{n} \\ \mathrm{n}+\mathrm{p}+\mathrm{m} & \mathrm{p} & \mathrm{m}\end{array}\right|$
Take $\mathrm{m}+\mathrm{n}+\mathrm{p}$ common from $\mathrm{C}_{1}$. $=(\mathrm{m}+\mathrm{n}+\mathrm{p})\left|\begin{array}{ccc}1 & \mathrm{n} & \mathrm{p} \\ 1 & \mathrm{~m} & \mathrm{n} \\ 1 & \mathrm{p} & \mathrm{m}\end{array}\right|$
$=(m+n+p)\left[\left(m^{2}+n^{2}+p^{2}\right)-m n-n p-p m\right]$
Hence, value of the determinant has linear factor.
$\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$
$=\left|\begin{array}{ccc}\mathrm{m}+\mathrm{n}+\mathrm{p} & \mathrm{n} & \mathrm{p} \\ \mathrm{p}+\mathrm{m}+\mathrm{n} & \mathrm{m} & \mathrm{n} \\ \mathrm{n}+\mathrm{p}+\mathrm{m} & \mathrm{p} & \mathrm{m}\end{array}\right|$
Take $\mathrm{m}+\mathrm{n}+\mathrm{p}$ common from $\mathrm{C}_{1}$. $=(\mathrm{m}+\mathrm{n}+\mathrm{p})\left|\begin{array}{ccc}1 & \mathrm{n} & \mathrm{p} \\ 1 & \mathrm{~m} & \mathrm{n} \\ 1 & \mathrm{p} & \mathrm{m}\end{array}\right|$
$=(m+n+p)\left[\left(m^{2}+n^{2}+p^{2}\right)-m n-n p-p m\right]$
Hence, value of the determinant has linear factor.
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