Prove that:

Solution:

Let $\Delta_{1}=\mid z \quad x \quad y$

$z^{2} \quad x^{2} \quad y^{2}$

$z^{4} \quad x^{4} \quad y^{4}\left|, \Delta_{2}=\right| x \quad y \quad z$

$x^{2} \quad y^{2} \quad z^{2}$

$x^{4} \quad y^{4} \quad z^{4}\left|, \Delta_{3}=\right| x^{2} \quad y^{2} \quad z^{2}$

$x^{4} \quad y^{4} \quad z^{4}$

$x \quad y \quad z \mid$ and $\Delta_{4}=x y z(x-y)(y-z)(z-x)(x+y+z)$

Now,

$\begin{array}{cccc}\Delta_{1}= & \mid z & x & y \\ z^{2} & x^{2} & y^{2} & \\ z^{4} & x^{4} & y^{4} \mid & \end{array}$

Using the property that if two rows ( or columns) of a determinant are interchanged, the value of the determinant becomes negetive, we get

$\Rightarrow \Delta_{1}=\left(\begin{array}{llll}-1 & \mid x & z & y\end{array}\right.$

$x^{2} \quad z^{2} \quad y^{2}$

$x^{4} \quad z^{4} \quad y^{4} \mid \quad\left[\because \mathbf{C}_{1} \leftrightarrow \mathbf{C}_{2}\right]$

$=(-1)(-1) \mid x \quad y \quad z$

$\begin{array}{ccc}x^{2} & y^{2} & z^{2} \\ x^{4} & y^{4} & z^{4}\end{array}$

$\mid \quad\left[\because C_{2} \leftrightarrow C_{3}\right]$

$=\mid \begin{array}{lll}\mathrm{x} & \mathrm{y} & \mathrm{z}\end{array}$

$\mathrm{x}^{2} \quad \mathrm{y}^{2} \quad \mathrm{z}^{2}$

$\begin{array}{lll}x^{4} & y^{4} & z^{4}\end{array}=\Delta_{2}$ $\cdots(1)$\

$=(-1) \mid x^{2} \quad y^{2} \quad z^{2}$

$\begin{array}{lll}x & y & z \\ x^{4} & y^{4} & z^{4} \mid\end{array}$ [Applying $\mathrm{R}_{1} \leftrightarrow \mathrm{R}_{2}$ ]

$=(-1)(-1) \mid \begin{array}{lll}\mathrm{x}^{2} & \mathrm{y}^{2} & \mathrm{z}^{2}\end{array}$

$\begin{array}{ccc}x^{4} & y^{4} & z^{4} \\ x & y & z \mid\end{array}$ [Applying $\mathrm{R}_{2} \leftrightarrow \mathrm{R}_{3}$ ]

$=\mid \begin{array}{lll}x^{2} & y^{2} & z^{2}\end{array}$

$\begin{array}{lll}x^{4} & y^{4} & z^{4} \\ x & y & z \mid=\Delta_{3}\end{array}$        ....(2)

Thus,

$\Delta_{1}=\Delta_{2}=\Delta_{3}$

[From eqs. (1) and (2)]