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Question

suppose a,b,cR and abc=1. If A=2abcb2caca2b is such that AA=41/3I and |A|>0, find the value of a3+b3+c3.

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Solution

Givenabc=1andA=2abcb2caca2bsuchthatATA=41/3Iand|A|>0

ATA=413IATA=(413)3|I|

|A|2=4|A|=2

|A|=∣ ∣2abcb2caca2b∣ ∣=10abc2(a3+b3+c3)=2

Hence,a3+b3+c3=4

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