#### Please Solve R.D.Sharma class 12 Chapter determinants Exercise  5.2 Question 35 Maths textbook Solution.

Answer:$4 a b c$

Hint Use determinant formula

Given : $\left|\begin{array}{ccc} \frac{a^{2}+b^{2}}{c} & c & c \\ a & \frac{b^{2}+c^{2}}{a} & a \\ b & b & \frac{c^{2}+a^{2}}{b} \end{array}\right|=4 a b c$

Solution:

$\text { L.H.S }\left|\begin{array}{ccc} \frac{a^{2}+b^{2}}{c} & c & c \\ a & \frac{b^{2}+c^{2}}{a} & a \\ b & b & \frac{c^{2}+a^{2}}{b} \end{array}\right|$

\begin{aligned} &R_{1} \rightarrow c R_{1}, R_{2} \rightarrow c R_{2}, R_{3} \rightarrow b R_{3} \\ &=\frac{1}{a b c}\left|\begin{array}{ccc} a^{2}+b^{2} & c^{2} & c^{2} \\ a^{2} & b^{2}+c^{2} & a^{2} \\ b^{2} & b^{2} & c^{2}+a^{2} \end{array}\right| \end{aligned}

\begin{aligned} &\text { Use } R_{1} \rightarrow R_{1}-R_{2}-R_{3} \\ &=\frac{1}{a b c}\left|\begin{array}{ccc} 0 & -2 b^{2} & -2 a^{2} \\ a^{2} & b^{2}+c^{2} & a^{2} \\ b^{2} & b^{2} & c^{2}+a^{2} \end{array}\right| \\ &\text { Now } \mathrm{C}_{2} \rightarrow C_{2}-C_{1} \text { and } \mathrm{C}_{3} \rightarrow C_{3}-C_{1} \\ &=\frac{1}{a b c} \mid \begin{array}{ccc} 0 & -2 b^{2} & -2 a^{2} \\ a^{2} & b^{2}+c^{2}-a^{2} & a^{2} \\ b^{2} & b^{2} & c^{2}+a^{2}-b^{2} \end{array} \end{aligned}

\begin{aligned} &\text { Expanding w.r.t } \mathrm{C}_{1}\\ &=\frac{1}{a b c}\left[c-a^{2}\left(-2 b^{2}\left(c^{2}+a^{2}-b^{2}\right)-a\right)+b^{2}\left(0-\left(2 a^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)\right)\right]\\ &=\frac{1}{a b c}\left[2 a^{2} b^{2}\left(c^{2}+a^{2}-b^{2}\right)+b^{2}\left(\left(2 a^{2}\right)\left(b^{2}+c^{2}-a^{2}\right)\right)\right]\\ &=\frac{2 a^{2} b^{2}}{a b c}\left(c^{2}+a^{2}-b^{2}+b^{2}+c^{2}-a^{2}\right)\\ &=\frac{2 a b}{c}\left(2 c^{2}\right)=4 a b c\\ &=R \cdot H \cdot S \end{aligned}