#### Provide Solution For  R.D. Sharma Maths Class 12 Chapter  determinants Exercise 5.2 Question 16 Maths Textbook Solution.

Answer:$\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|=(a-b)(b-c)(c-a)$

Hint : We will try to make some elements of the determinant zero

Given:$\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|=(a-b)(b-c)(c-a)$

Solution:

$\text { L.H.S }\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|$

If the rows and columns of a determinant are interchanged, the value of the determinant remains the same.

$=\left|\begin{array}{ccc} 1 & 1 & 1 \\ b+c & c+a & a+b \\ b^{2}+c^{2} & c^{2}+a^{2} & a^{2}+b^{2} \end{array}\right|$

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

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

\begin{aligned} &\text { On taking common }(b-a) \text { from } \mathrm{C}_{1} \text { and }(c-b) \text { from } \mathrm{C}_{2}\\ &=(b-a)(c-b)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & 1 & a+b \\ (b+a) & (c+b) & a^{2}+b^{2} \end{array}\right| \end{aligned}

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

Hence it is proved that

$\left|\begin{array}{lll} 1 & b+c & b^{2}+c^{2} \\ 1 & c+a & c^{2}+a^{2} \\ 1 & a+b & a^{2}+b^{2} \end{array}\right|=(a-b)(b-c)(c-a)$