Groups of Reciprocal Friends

$\small {\color{#3D99F6}{\text{Clearly roots of }x^2+x+1 \text{are }ω,ω^2 \text{such that}~ ω^3=1}}$ $\color{forestgreen}{\bullet~ω=ω^4=ω^7=ω^{10}=... \\ {\bullet} ~ω^2=ω^5=ω^8=ω^{11}=... \\ {\bullet}~ ω^3=ω^6=ω^9=ω^{12}=....=1}$ Using cyclicity of ω as stated above, required sum (which has the same value at ω as well as $ω^2$ ) simplifies to: $9( ({x+\frac { 1 }{ x } ) }^{ 2 }+({ { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ) }^{ 2 }+{ ({ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } ) }^{ 2 })$ $9( (ω+ω^2)^2+(ω^2+ω)^2+(1+1)^2$ $~~~~~\small{\color{#D61F06}{(\dfrac{1}{ω}=ω^2)}}$ $=9\times((-1)^2+(-1)^2+4)=\Large 54$