Let `omega` be the solution of `x^(3)-1=0` with `"Im"(omega) gt 0`. If a=2 with b and c satisfying `[abc][{:(1,9,7),(2,8,7),(7,3,7):}]=[0,0,0]`, then the value of `3/omega^(a) + 1/omega^(b) + 1/omega^( c)` is equal to

Options

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

Answer

A

Explanation

We have,
`[abc][{:(1,9,7),(2,8,7),(7,3,7):}]=[0,0,0]`
`rArr [a+2b+7c, 9a+8b+3c, 7a+7b+7c]=[0 0 0]`
`rArr a+2b+7c=0, 9a+8b+3c=0`and `7a+7b+7c=0`
`rArr a=5,b=-6,c=1`
Clearly, `omega=-1/sqrt(2)+isqrt(3)/2` and it satisfies `omega^(3)=1, 1/omega=omega^(2)`
`therefore 3/omega^(a)+1/omega^(b)+3/omega^(c)=3/omega^(5)+1/omega^(-6)+3/omega^(1)=3omega+1+3omega^(2)`
`=3(omega^(2)+omega)+1=-3+1=-2`