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

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

To 2 decimal places, find the value of -

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

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

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