If x+1z=1, then z20+1z40+1z60+1z60=

z+1z=1 ⇒ z2-z+1=0∴ z=-w or -w2z20+1 z40+1 z60+1z60=w20+1 w40+1 w60+1w60 =w2+1 w+1 1+11 =2 w3+w2+w+1 =22-1 =2

If x+1z=1, then z20+1z40+1z60+1z60=

z+1z=1 ⇒ z2-z+1=0∴ z=-w or -w2z20+1 z40+1 z60+1z60=w20+1 w40+1 w60+1w60 =w2+1 w+1 1+11 =2 w3+w2+w+1 =22-1 =2

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If $x + \frac{1}{z} = 1,$ then $\frac{(z^{20} + 1) (z^{40} + 1) (z^{60} + 1)}{z^{60}} =$

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Detailed Solution

$\begin{array}{rcl} z + \frac{1}{z} & = & 1 \Rightarrow z^{2} - z + 1 = 0 \end{array}$

$\begin{array}{rcl} ∴ z & = & - w or - w^{2} \end{array}$

$\begin{array}{rcl} \frac{(z^{20} + 1) (z^{40} + 1) (z^{60} + 1)}{z^{60}} & = & \frac{(w^{20} + 1) (w^{40} + 1) (w^{60} + 1)}{w^{60}} \\ = & \frac{(w^{2} + 1) (w + 1) (1 + 1)}{1} \\ = & 2 (w^{3} + w^{2} + w + 1) \\ = & 2 (2 - 1) \\ = & 2 \end{array}$