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
`a=2, b` and c satisfies (E)
`b=12, c=-14`
`3/omega^(a)+1/omega^(b)+3/omega^(c)`
`=3/omega^(2)+1/omega^(12) +3/omega^(-14)`
`=3omega+1+3 omega^(2)`
`=-2`.