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Question: \[\left| \begin{matrix} b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{matrix} \ri...

b + c & a - b & a \\ c + a & b - c & b \\ a + b & c - a & c \end{matrix} \right| =$$
A

a3+b3+c3−3abca^{3} + b^{3} + c^{3} - 3abc

B

3abc−a3−b3−c33abc - a^{3} - b^{3} - c^{3}

C

a3+b3+c3−a2b−b2c−c2aa^{3} + b^{3} + c^{3} - a^{2}b - b^{2}c - c^{2}a

D

3abc−a3−b3−c33abc - a^{3} - b^{3} - c^{3}

Answer

3abc−a3−b3−c33abc - a^{3} - b^{3} - c^{3}

Explanation

Solution

Δ=∣2(a+b+c)0a+b+cc+ab−cba+bc−ac∣\Delta = \left| \begin{matrix} 2(a + b + c) & 0 & a + b + c \\ c + a & b - c & b \\ a + b & c - a & c \end{matrix} \right|

by R1→R1+R2+R3R_{1} \rightarrow R_{1} + R_{2} + R_{3}

2 & 0 & 1 \\ c + a & b - c & b \\ a + b & c - a & c \end{matrix} \right|$$ On expanding, $- (a + b + c)(a^{2} + b^{2} + c^{2} - ab - bc - ca)$ = $= a_{1}^{2}(b_{2}c_{3} - b_{3}c_{2}) + a_{1}b_{1}( - c_{3}a_{2} + a_{3}c_{2})$. **Trick :** Put $a = 1,b = 2,c = 3$ and check it.