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Question

Let a,b,cR be such that a+b+c>0 and abc=2. Let A=abcbcacab If A2=I, then value of a3+b3+c3 is

A
7
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B
2
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C
0
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D
-1
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Solution

The correct option is D 7
A2=I
a2+b2+c2=1 and bc+ca+ab=0
Also, |A2|=1|A|=±1.
But A|=3abc(a3+b3+c3)
=(a+b+c)(a2+b2+c2bccaab)
=(a+b+c)
As a+b+c>0, we get |A|=1.
a3+b3+c3=7

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