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Need Solution for R. D.  Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 2 sub question 12 Maths Textbook Solution.

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Answer:\left|\begin{array}{lll} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right|=3 a b c-a^{3}-b^{3}-c^{3}

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

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

Solution:L.H.S\left|\begin{array}{lll} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right|

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

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(a+b+c)\left[2\left|\begin{array}{ll} b-c & b \\ c-a & c \end{array}\right|-0\left|\begin{array}{ll} c+a & b \\ a+b & c \end{array}\right|+1\left|\begin{array}{ll} c+a & b-c \\ a+b & c-a \end{array}\right|\right.\\ &=(a+b+c)[1\{(b-c)(c)-(b)(c-a)\}-0+1\{(c+a)(c-a)-(b-c)(a+b)\}]\\ &=(a+b+c)\left[2\left(b c-c^{2}-b c+a b\right)+1\left\{c^{2}-a^{2}-\left(a b-a c+b^{2}-b c\right)\right\}\right]\\ &=(a+b+c)\left\{2\left(-c^{2}+a b\right)+1\left(c^{2}-a^{2}-a b+a c-b^{2}+b c\right)\right\}\\ &=(a+b+c)\left(-2 c^{2}+2 a b+c^{2}-a^{2}-a b+a c-b^{2}+b c\right)\\ &=(a+b+c)\left(c^{2}-a^{2}-b^{2}-a b+a c+b c\right)\\ &=-(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\\ &=-\left(a^{3}+b^{3}+c^{3}-3 a b c\right)\\ &=-a^{3}-b^{3}-c^{3}+3 a b c \quad \because x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)\\ &=3 a b c-a^{3}-b^{3}-c^{3} \end{aligned}

=RHS

Hence it is proved that

\left|\begin{array}{lll} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{array}\right|=3 a b c-a^{3}-b^{3}-c^{3}

 

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