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Note that $\displaystyle 4^{x} + 4^{-x} = 2^{2x} + 2^{-2x}$ And we know that $\displaystyle 2^{x} + 2^{-x} = 3$. By squaring both sides we get: $\displaystyle 2^{2x} + 2(2^{0}) + 2^{-2x} = 9$
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