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)]

$\Delta_{2}=\mid x \quad y \quad z$

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

$=x y z \mid \begin{array}{lll}1 & 1 & 1\end{array}$

$\mathrm{x}^{3} \quad \mathrm{y}^{3} \quad \mathrm{z}^{3} \mid \quad\left[\right.$ Taking out common factor $\mathrm{x}$ from $\mathrm{C}_{1}, \mathrm{y}$ from $\mathrm{C}_{2}$ and $\mathrm{z}$ from $\left.\mathrm{C}_{3}\right]$

$=x y z \mid 0 \quad 0 \quad 1$

$\begin{array}{lll}x-y & y-z & z\end{array}$

$x^{3}-y^{3} \quad y^{3}-z^{3} \quad z^{3} \mid \quad$ [Applying $\mathrm{C} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}$ and $\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3}$ ]

$=\mathrm{xyz}(x-y)(y-z) \mid \quad 0 \quad 0 \quad 1 \quad 1 \quad$ $\quad \begin{array}{lllll} & 0 & & z x^{2}+2 x y+y^{2} & y^{2}+2 y z+z^{2} & z^{3} \mid\end{array}$

$\left[\because\left(a^{3}-b^{3}\right)=(a-b)\left(a^{2}+a b+b^{2}\right)\right] \quad\left[\right.$ Taking out common factor $(x-y)$ from $C_{1}$ and $(y-z)$ from $\left.C_{2}\right]$

$=\mathrm{xyz}(x-y)(y-z)\left\{1 \times\left|1 \quad 1 \quad x^{2}+x y+y^{2} \quad y^{2}+y z+z^{2}\right|\right\} \quad\left[\right.$ Expanding along $\left.R_{1}\right]$

$=\mathrm{xyz}(x-y) \quad(y-z)\left\{y^{2}+y z+z^{2}-x^{2}-x y-y^{2}\right\}$

$=\mathrm{xyz}(x-y) \quad(y-z)\left\{y z-x y+z^{2}-x^{2}\right\}$

$=\mathrm{xyz}(x-y) \quad(y-z)\{y(z-x)+(z-x)(z+x)\}$

$=\mathrm{xyz}(x-y) \quad(y-z)(z-x)(y+x+z)$

$=\mathrm{xyz}(x-y) \quad(y-z)(z-x)(x+y+z)$

$=\Delta_{4}$

Thus,

$\Delta_{1}=\Delta_{2}=\Delta_{3}=\Delta_{4}$