#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 43 Maths Textbook Solution.

Answer:$(x+y+z)(x-z)^{2}$

Hint Use determinant formula

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

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} y+z & x & y \\ z+x & z & x \\ x+y & y & z \end{array}\right| \\ &\text { Apply } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\left|\begin{array}{ccc} 2(x+y+z) & x+y+z & x+y+z \\ z+x & z & x \\ x+y & y & z \end{array}\right| \end{aligned}

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

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

\begin{aligned} &=(x+y+z)\left((z-x)^{2}-0\right) \\ &=(x+y+z)(x-z)^{2} \\ &=R \cdot H \cdot S \end{aligned}