#### Need Solution for R.D. Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question  20 Maths Textbook Solution.

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

Hint We will try to make $\left ( a-b \right )$and $\left ( b-c \right )$any two elements of determinant

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

Solution:$\text { L.H.S }\left|\begin{array}{lll} (b+c)^{2} & a^{2} & b c \\ (c+a)^{2} & b^{2} & c a \\ (a+b)^{2} & c^{2} & a b \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3}\\ &=\left|\begin{array}{ccc} (b+c)^{2}-(c+a)^{2} & a^{2}-b^{2} & b c-c a \\ (c+a)^{2}-(a+b)^{2} & b^{2}-c^{2} & c a-a b \\ (a+b)^{2} & c^{2} & a b \end{array}\right|\\ &x^{2}-y^{2}=(x+y)(x-y) \end{aligned}

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

\begin{aligned} &\text { Taking common }(a-b) \text { from } \mathrm{R}_{1},(b-c) \text { from } \mathrm{R}_{2}\\ &=(a-b)(b-c)\left|\begin{array}{ccc} -(a+b+2 c) & (a+b) & -c \\ -(2 a+b+c) & (b+c) & -a \\ (a+b)^{2} & c^{2} & a b \end{array}\right|\\ &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2}\\ &=(a-b)(b-c)\left|\begin{array}{ccc} -a-b-2 c+2 a+b+c & a+b-b-c & -c+a \\ -(2 a+b+c) & (b+c) & -a \\ (a+b)^{2} & c^{2} & a b \end{array}\right| \end{aligned}

\begin{aligned} &=(a-b)(b-c)\left|\begin{array}{ccc} a-c & a-c & a-c \\ -(2 a+b+c) & (b+c) & -a \\ (a+b)^{2} & c^{2} & a b \end{array}\right|\\ &\text { Taking common }(a-c) \text { from } \mathrm{R}_{1}\\ &=(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 1 & 1 & 1 \\ -(2 a+b+c) & (b+c) & -a \\ (a+b)^{2} & c^{2} & a b \end{array}\right| \end{aligned}

\begin{aligned} &\text { On applying } C_{1} \rightarrow C_{1}-C_{2} \text { and } C_{2} \rightarrow C_{2}-C_{3}\\ &=(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -2 a-b-c-b-c & b+c+a & -a \\ (a+b)^{2}-c^{2} & c^{2}-a b & a b \end{array}\right|\\ &=(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -2 a-2 b-2 c & a+b+c & -a \\ (a+b+c)(a+b-c) & c^{2}-a b & a b \end{array}\right|\\ &=(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -2(a+b+c) & a+b+c & -a \\ (a+b+c)(a+b-c) & c^{2}-a b & a b \end{array}\right| \end{aligned}

Taking common(a+b+c)from$C_{1}$

\begin{aligned} &=(a+b+c)(a-b)(b-c)(a-c)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -2 & a+b+c & -a \\ (a+b-c) & c^{2}-a b & a b \end{array}\right| \\ &\quad=-(a+b+c)(a-b)(b-c)(c-a)\left|\begin{array}{ccc} 0 & 0 & 1 \\ (a+b-c) & c^{2}-a b & a b \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left[0\left|\begin{array}{cc} a+b+c & -a \\ c^{2}-a b & a b \end{array}\right|-0\left|\begin{array}{cc} -2 & -a \\ a+b-c & a b \end{array}\right|\right.\\ &\left.+1\left|\begin{array}{cc} -2 & a+b+c \\ a+b-c & c^{2}-a b \end{array}\right|\right]\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left[0-0+1\left\{-2\left(c^{2}-a b\right)-(a+b-c)(a+b+c)\right\}\right]\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left[-2 c^{2}+2 a b-\left\{(a+b)^{2}-c^{2}\right\}\right]\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left[-2 c^{2}+2 a b-\left\{a^{2}+b^{2}+2 a b-c^{2}\right\}\right]\\ &\because(a+b)^{2}=a^{2}+b^{2}+2 a b\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left(-2 c^{2}+2 a b-a^{2}-b^{2}-2 a b-c^{2}\right)\\ &=-(a+b+c)(a-b)(b-c)(c-a)\left(-a^{2}-b^{2}-c^{2}\right) \end{aligned}

\begin{aligned} &=(-)(-)(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \\ &=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \end{aligned}

Hence it is proved that

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