$a^{4x}=\sqrt{2}-1$ $\ \ \ \frac{1}{a^{4x}}=\sqrt{2}+1$
$$\frac{a^{6x}+a^{-6x}}{a^{2x}+a^{-2x}}=\frac{a^{12x}+1}{a^{8x}+a^{4x}}$$
$$=\frac{(a^{4x}+1)(a^{8x}-a^{4x}+1)}{a^{4x}(a^{4x+1})}$$
$$=a^{4x}-1+\frac{1}{a^{4x}}=(\sqrt{2}-1)-1+(\sqrt{2}+1)=2\sqrt{2}-1$$
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