Too Complex! Part- II

Algebra Level 3

If Real numbers a, b, c and d (none of which are 1 -1 ) satisfy the following conditions:

1a+ω+1b+ω+1c+ω+1d+ω=2ω2, \frac{1}{a+ ω} + \frac{1}{b+ ω}+ \frac{1}{c+ ω} + \frac{1}{d+ ω} = 2 ω^{2}, 1a+ω2+1b+ω2+1c+ω2+1d+ω2=2ω. \frac{1}{a+ ω^{2}}+ \frac{1}{b+ ω^{2}} + \frac{1}{c+ ω^{2}} + \frac{1}{d+ ω^{2}} = 2 ω.

To 2 decimal places, find the value of -

1a+1+1b+1+1c+1+1d+1 \frac{1}{a+ 1} + \frac{1}{b+ 1}+ \frac{1}{c+ 1} + \frac{1}{d+ 1}


Note that ω and ω2ω^{2} are the complex cube roots of unity other than 1.

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