#### Explain Solution R. D. Sharma Class 12 Chapter  deteminants Exercise 5.2 Question 42  maths Textbook Solution.

Answer: $(x+y+z)^{3}$

Hint Use determinant formula

Given:$\left|\begin{array}{ccc} 2 y & y-z-x & 2 y \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right|=(x+y+z)^{3}$

$\text { L.H.S }\left|\begin{array}{ccc} 2 y & y-z-x & 2 y \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right|$

Solution:

\begin{aligned} &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} x+y+z & x+y+z & x+y+z \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right| \\ &(x+y+z) \text { common from } \mathrm{C}_{1} \\ &=(x+y+z)\left|\begin{array}{ccc} 1 & 1 & 1 \\ 2 z & 2 z & z-x-y \\ x-y-z & 2 x & 2 x \end{array}\right| \end{aligned}

\begin{aligned} &\text { Apply } C_{2} \rightarrow C_{2}-C_{1} \& \mathrm{C}_{3} \rightarrow C_{3}-C_{1} \\ &=(x+y+z)\left|\begin{array}{ccc} 1 & 0 & 1 \\ 2 z & 0 & -(x+y+z) \\ x-y-z & x+y+z \quad & x+y+z \end{array}\right| \\ &(x+y+z) \text { common from } \mathrm{C}_{2} \& C_{3} \\ &=(x+y+z)^{3}\left|\begin{array}{ccc} 1 & 0 & 1 \\ 2 z & 0 & -1 \\ x-y-z & 1 & 1 \end{array}\right| \end{aligned}

Expanding w.r.t $C_{3}$

$=(x+y+z)^{3}\left[1\left|\begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array}\right|-0(1+0) 1\right]$

\begin{aligned} &=(x+y+z)^{3}\left|\begin{array}{cc} 0 & -1 \\ 1 & 1 \end{array}\right| \\ &=(x+y+z)^{3}(0(1)-(-1) 1) \\ &=(x+y+z)^{3} \\ &=R \cdot H . S \end{aligned}