#### Provide Solution For  R. D. Sharma Maths Class 12 Chapter  determinants Exercise 5.2 Question 31 Maths Textbook Solution.

Answer: $2(a+b)(b+c)(c+a)$

Hint Use determinant formula

Given:$\left|\begin{array}{ccc} a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{array}\right|=2(a+b)(b+c)(c+a)$

Solution:

\begin{aligned} &\text { L.H.S }\left|\begin{array}{ccc} a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{array}\right| \\ &\text { Use } \mathrm{C}_{1} \rightarrow C_{1}+\mathrm{C}_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}+\mathrm{C}_{3} \\ &|\Delta|=\left|\begin{array}{ccc} a+b & -b-c & -b \\ a+b & b+c & -a \\ -a-b & b+c & a+b+c \end{array}\right| \end{aligned}

\begin{aligned} &\text { Common }(a+b)(b+c) \text { from } \mathrm{C}_{1} \& C_{2} \\ &|\Delta|=(a+b)(b+c)\left|\begin{array}{ccc} 1 & -1 & -b \\ 1 & 1 & -a \\ -1 & 1 & a+b+c \end{array}\right| \\ &\text { Use } \mathrm{R}_{1} \rightarrow R_{1}+\mathrm{R}_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}+\mathrm{R}_{3} \\ &|\Delta|=(a+b)(b+c)\left|\begin{array}{ccc} 2 & 0 & -b+a \\ 0 & 2 & b+c \\ -1 & 1 & a+b+c \end{array}\right| \end{aligned}

\begin{aligned} &\text { Expanding w.r.t } \mathrm{C}_{1}\\ &|\Delta|=(a+b)(b+c)[2(2 a+2 b+2 c-b-c)+(-1)(2 b+2 a)]\\ &|\Delta|=(a+b)(b+c)[4 a+4 b+4 c-2 b-2 c-2 b-2 a]\\ &|\Delta|=2(a+b)(b+c)(c+a) \end{aligned}