If `A=[(a,b,c),(b,c,a),(c,a,b)],abc=1,A^(T)A=I,` then find the value of `a^(3)+b^(3)+c^(3).`

Answer

9

Explanation

`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`