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Question

∣ ∣b+cabac+abcba+bcac∣ ∣=


A
a3+b3+c33abc
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B
3abca3b3c3
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C
a3+b3+c3a2bb2cc2a
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D
(a+b+c)(a2bb2cc2+ab+bc+ca)
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Solution

The correct option is A 3abca3b3c3
Δ=∣ ∣2(a+b+c)0a+b+cc+abcba+bcac∣ ∣by R1R1+R2+R3Δ=(a+b+c)∣ ∣201c+abcba+bcac∣ ∣
On expanding, (a+b+c)(a2+b2+c2abbcca)
=(a3+b3+c33abc)=3abca3b3c3.
Trick : Put a=1, b=2, c=3 and check it.
 

Mathematics

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