Solution
$$x+\ \frac{\ 1}{16x}=1$$
Multiplying by 4 on both sides we get
$$4x+\ \frac{\ 1}{4x}=4$$
$$\left(4x+\ \frac{\ 1}{4x}\right)^3=4^3$$
$$64x^3\ +\ \ \frac{\ 1}{64x^3}+3.4x.\ \frac{\ 1}{4x}\left(4x+\ \frac{\ 1}{4x}\right)=64$$
$$64x^3\ +\ \ \frac{\ 1}{64x^3}+3\left(4\right)=64$$
$$64x^3\ +\ \ \frac{\ 1}{64x^3}=52$$
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