Question

Let a,b, and c be three real numbers satistying `[a,b,c][(1,9,7),(8,2,7),(7,3,7)]=[0,0,0]` Let `omega` be a solution of `x^3-1=0` with `Im(omega)gt0. I fa=2` with b nd c satisfying (E) then the vlaue of `3/omega^a+1/omega^b+3/omega^c` is equa to (A) -2 (B) 2 (C) 3 (D) -3
A. `-2`
B. 2
C. 3
D. `-3`

Answer 1

Correct Answer - A
If a = 2, b= 12, c=-14
`therefore (3)/(omega^(a)) + (1)/(omega^(b)) + (3)/(omega^(c))`
`rArr (3)/(omega^(2)) + (1)/(omega^(12)) + (3)/(omega^(-14)) = (3)/(omega^(2)) + 1 + 3omega^(2) = 3omega + 1 +3omega^(2)`
` = 1+ 3 (omega + omega^(2)) = 1-3 =-2`