By using properties of determinants, show that:
$\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$
By using properties of determinants, show that:
$\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|=(x-y)(y-z)(z-x)(x y+y z+z x)$
Let $\Delta=\left|\begin{array}{lll}x & x^{2} & y z \\ y & y^{2} & z x \\ z & z^{2} & x y\end{array}\right|$
Applying $R_{2} \rightarrow R_{2}-R_{1},$ and $R_{3} \rightarrow R_{3}-R_{1},$ we have:
$\Delta=\left|\begin{array}{ccc}x & x^{2} & y z \\ y-x & y^{2}-x^{2} & z x-y z \\ z-x & z^{2}-x^{2} & x y-y z\end{array}\right|$
$=\left|\begin{array}{ccc}x & x^{2} & y z \\ -(x-y) & -(x-y)(x+y) & z(x-y) \\ (z-x) & (z-x)(z+x) & -y(z-x)\end{array}\right|$
$=(x-y)(z-x)\left|\begin{array}{ccc}x & x^{2} & y z \\ -1 & -x-y & z \\ 1 & z-y & z-y\end{array}\right|$
Applying $R_{3} \rightarrow R_{3}+R_{2},$ we have:
$\Delta=(x-y)(z-x)\left|\begin{array}{ccc}x & x^{2} & y z \\ -1 & -x-y & z \\ 1 & z-y & z-y\end{array}\right|$
$=(x-y)(z-x)(z-y)\left|\begin{array}{ccc}x & x^{2} & y z \\ -1 & -x-y & z \\ 0 & 1 & 1\end{array}\right|$
Expanding along $R_{3},$ we have:
$\Delta=[(x-y)(z-x)(z-y)]\left[(-1)\left|\begin{array}{cc}x & y z \\ -1 & z\end{array}\right|+1\left|\begin{array}{cc}x & x^{2} \\ -1 & -x-y\end{array}\right|\right]$
$=(x-y)(z-x)(z-y)\left[(-x z-y z)+\left(-x^{2}-x y+x^{2}\right)\right]$
$=-(x-y)(z-x)(z-y)(x y+y z+z x)$
$=(x-y)(y-z)(z-x)(x y+y z+z x)$
Hence, the given result is proved.
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If $\left| {\begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}} \\
{{{(a + \lambda )}^2}}&{{{(b + \lambda )}^2}}&{{{(c + \lambda )}^2}} \\
{{{(a - \lambda )}^2}}&{{{(b - \lambda )}^2}}&{{{(c - \lambda )}^2}}
\end{array}} \right|$ $ = \,k\lambda \,\,\left| {{\mkern 1mu} {\mkern 1mu} \begin{array}{*{20}{c}}
{{a^2}}&{{b^2}}&{{c^2}} \\
a&b&c \\
1&1&1
\end{array}} \right|,\lambda \, \ne \,0$ then $k$ is equal to
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{{a^2}}&{{b^2}}&{{c^2}} \\
{{{(a + \lambda )}^2}}&{{{(b + \lambda )}^2}}&{{{(c + \lambda )}^2}} \\
{{{(a - \lambda )}^2}}&{{{(b - \lambda )}^2}}&{{{(c - \lambda )}^2}}
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