Show that: y+z&x&y +x&z&x +y&y&z =(x+y+z)(x-z)^3 - Vidyadip

Question

Show that:
$\begin{vmatrix}\text{y}+\text{z}&\text{x}&\text{y}\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix}=(\text{x}+\text{y}+\text{z})(\text{x}-\text{z})^3$

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Answer

Let $\text{L.H.S}=\begin{vmatrix}\text{y}+\text{z}&\text{x}&\text{y}\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix}$
$=\begin{vmatrix}2(\text{x}+\text{y}+\text{z})&\text{x}+\text{y}+\text{z}&\text{x}+\text{y}+\text{z}\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix}$ [Applying $R_1 \rightarrow R_1 + R_2 + R_3]$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}2&1&1\\\text{z}+\text{x}&\text{z}&\text{x}\\\text{x}+\text{y}&\text{y}&\text{z}\end{vmatrix}$
$=(\text{x}+\text{y}+\text{z})\begin{vmatrix}0&1&1\\0&\text{z}&\text{x}\\\text{x}-\text{z}&\text{y}&\text{z}\end{vmatrix}$ [Applying $C_1 \rightarrow C_1 - C_2 - C_3]$
$=(\text{x}+\text{y}+\text{z})\left\{(\text{x}-\text{z})\times\begin{vmatrix}1&1\\\text{z}&\text{x}\end{vmatrix}\right\}$ [Expanding along $C_1]$
$=(\text{x}+\text{y}+\text{z})(\text{x}-\text{z})^3$

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