Complex City

${ x }^{ 6 }-2{ x }^{ 3 }+1=0\\ { ({ x }^{ 3 }-1) }^{ 2 }=0\\ { x }^{ 3 }=1\\ \\ \\ \\ \sum _{ r=1 }^{ 50 }{ { ({ x }^{ r }+{ x }^{ 2r } })^{ 3 } } =\sum _{ r=1 }^{ 50 }{ { { x }^{ 3r }(1+{ x }^{ r } })^{ 3 } } \\ \\ because\quad { x }^{ 3 }=1,\quad \sum _{ r=1 }^{ 50 }{ { { x }^{ 3r }(1+{ x }^{ r } })^{ 3 } } =\sum _{ r=1 }^{ 50 }{ { (1+{ x }^{ r } })^{ 3 } } \\ \\ =\sum _{ r=1 }^{ 50 }{ 1+3 } { x }^{ r }+3{ x }^{ 2r }+{ x }^{ 3r }=\sum _{ r=1 }^{ 50 }{ 1 } +\sum _{ r=1 }^{ 50 }{ 3 } { x }^{ r }+\sum _{ r=1 }^{ 50 }{ 3{ x }^{ 2r } } +\sum _{ r=1 }^{ 50 }{ 1 } \quad \\ \\ =\quad 50+\quad 3\frac { x({ x }^{ 50 }-1) }{ x-1 } +\quad 3\frac { { x }^{ 2 }({ x }^{ 100 }-1) }{ { x }^{ 2 }-1 } +\quad 50\quad =50+\quad 3\times \frac { { x }^{ 51 }-x }{ x-1 } +\quad 3\times \frac { { x }^{ 102 }-{ x }^{ 2 } }{ { x }^{ 2 }-1 } +\quad 50\\ \\ =\quad 50+\quad 3\times \frac { 1-x }{ x-1 } +\quad 3\times \frac { 1-{ x }^{ 2 } }{ { x }^{ 2 }-1 } +\quad 50\quad \Leftarrow because\quad { x }^{ 3 }=1\\ \\ =50-3-3+50=94\\$

First we rewrite the equation as $(x^3 - 1)^2 = 0$ . This implies that $x^3 = 1$ . Because $x$ is not real, $x$ must be one of the cube roots of unity: $x = \omega\ \text{or}\ \bar\omega,$ where $\omega = e^{2\pi i/3} = -\tfrac12 \pm \tfrac12\sqrt 3 i$ . Note that $\omega^2 = \bar\omega$ and vice versa.

Now if $r = 3s$ is a multiple of 3, then $x^r + x^{2r} = (\omega^3)^s + (\bar\omega^3)^s = 1^s + 1^s = 2$ .

But if $r$ is not a multiple of 3, then $x^r + x^{2r} = \omega + \bar\omega = 2\text{Re}\ \omega = -1$ .

In the range $r = 1\dots50$ , we have 16 terms where $r$ is a multiple of 3, and 34 other terms. Thus the answer is $16\cdot 2^3 + 34\cdot (-1)^3 = 128 - 34 = \boxed{94}.$

Use the below thing to get the answer $x^{3}=1; x=ω,ω^{2}.$ $x^{r}+x^{2r}=2 ;$ if r is a multiple of 3. $x^{r}+x^{2r}=-1;$ if r is not a multiple of 3