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

Options

A. -2
B. 2
C. 3
D. -3

Answer

A

Explanation

`because a = 2 ` with b and c satisfying ( E)
`therefore 2+ 8b+ 7c = 0, 18 + 2b + 3c = 0`
and ` 2+ b + c= 0`
we get ` b = 12 and c= -14`
Hence, `3/omega^(a) + 1/omega^(b) + 3/omega^(c) = 3/omega^(2) + 1/omega^(12) + 3/omega^(-14) `
`= (3omega)/omega^(3) + 1/1 + 3omega^(14) `
`= 3 omega + 1 + 3omega^(2)`
`= 1 + 3(omega+omega^(2))`
`= 1+ 3(-1)=-2`