#### Please Solve R. D. Sharma class 12 Chapter determinants Exercise 5.2 Question 2 sub question 6 Maths textbook Solution.

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

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

Given :$\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|$

Solution: $\left|\begin{array}{lll} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|$

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

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

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

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