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Suppose a, b, c, in R and abc = 1, if A...

Suppose `a, b, c, in R ` and `abc = 1, if A = [[3a, b, c ],[b, 3c, a ],[c, a, 3b]]` is such that `A ^(T) A = 4 ^(1//3) I and abs(A) gt 0, ` the value of `a^(3) + b^(3) + c^(3)` is

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The correct Answer is:
9
`because A = [[3a, b, c],[c,3c, a],[c, a, 3b]]`
`therefore det (A)= [[3a, b, c],[c,3c, a],[c, a, 3b]]= 29 abc - 3(a^(3) + b^(3) +c^(3) )`
or
`abs(A)= 29 abc - 3(a^(3) + b^(3) +c^(3) ) " " ...(i)`
Given, `A^(T) A + 4^(1//3)I`
`rArr abs(A^(T)A) =abs( 4^(1//3)I)`
`rArr abs(A^(T))abs(A) = (4^(1//3))^(3)abs(I)`
`rArr abs(A)abs(A) = 4.1`
`therefore abs(A) = 2 " " [because abs(A) gt0]`
From Eq.(i) we get
`2 = 29abc - 3 ( a^(3) + b^(3) + c^(3) )`
`rArr 2 = 29 - 3 ( a^(3) + b^(3) + c^(3) ) " " because abc = 1]`
` therefore a^(3) + b^(3) c^(3) = 9`
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