ALgebra#3

$Call\ a^n =x,\ b^n=y\ and\ c^n=z.\ Then\ xy+yz+xz=138\ and\ x^2y^2+y^2z^2=543.\\ Try\ to\ expand\ (xy+xz+yz)^2\ and\ we\ obtain\ x^2y^2+y^2z^2+x^2z^2+2(xy^2+x^2yz+xyz^2)= x^2y^2+y^2z^2+x^2z^2+2xyz(y^2+x^2+z^2)\\ and\ we\ want\ this\ without\ the\ first\ two\ terms.\ Now\ x^2z^2+2xyz(y^2+x^2+z^2)= (xy+xz+yz)^2 -x^2y^2+y^2z^2= 138^2 - 543 =18501$

Let $\alpha = a^nb^n$ , $\beta = b^nc^n$ and $\gamma = c^na^n$ . Then we have:

$\begin{cases} a^nb^n + b^nc^n + c^na^n =\color{#3D99F6} \alpha + \beta + \gamma = 138 \\ a^{2n}b^{2n} + b^{2n}c^{2n} =\color{#D61F06} \alpha^2 + \beta^2 = 543 \\ c^{2n}a^{2n} + 2a^nb^nc^n(a^n+b^n + c^n) = \gamma^2 + 2(\gamma \alpha + \alpha \beta + \beta \gamma) \end{cases}$

We note that

$\begin{aligned} ({\color{#3D99F6}\alpha+\beta+\gamma})^2 & = {\color{#D61F06}\alpha^2+\beta^2}+\gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha) \\ {\color{#3D99F6}138}^2 & = {\color{#D61F06}543}+\gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha) \\ \implies \gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha) & = 138^2 - 543 = \boxed{18501} \end{aligned}$