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Question

b2c2bcb+cc2a2cac+aa2b2aba+b is equal to

A
1abc(ab+bc+ca)
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B
ab+bc+ca
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C
0
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D
a+b+c
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Solution

The correct option is A 0
b2c2bcb+cc2a2cac+aa2b2aba+b
R1a2R1,R2b2R2,R3c2R3
1a2b2c2⎢ ⎢a2b2c2a2bca2(b+c)b2c2a2b2cab2(c+a)a2b2c2c2abc2(a+b)⎥ ⎥
abc⎢ ⎢1aa2(b+c)1bb2(c+a)1cc2(a+b)⎥ ⎥
R1R1R2R2R2R3
abc⎢ ⎢0aba(b+c)b2(c+a)0bcb2(c+a)c2(a+b)1cc2(a+b)⎥ ⎥
abc0ab(ab)(ab+bc+ca)0bc(bc)(ab+bc+ca)1cc2(a+b)
(abc)(ab)(bc)01(ab+bc+ca)01(ab+bc+ca)1cc2(a+b)
On expanding the determinant, we have
(abc)(ab)(bc)(ab+bc+caabbcca)=0

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