$$\frac{1}{x}+8+\frac{16}{x^2}-64 = 1$$
Answer
$$ \begin{matrix}x_1 = 0.73611 & x_2 = -0.71956 & x_3 = -0.00828+0.72793i \\[1 em] x_4 = -0.00828-0.72793i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{1}{x}+8+\frac{16}{x^2}-64 &= 1&& \text{multiply ALL terms by } \color{blue}{ xx^2 }. \\[1 em]xx^2\cdot\frac{1}{x}+xx^2\cdot8+xx^2\cdot\frac{16}{x^2}-xx^2\cdot64 &= xx^2\cdot1&& \text{cancel out the denominators} \\[1 em]1+8x^3+\frac{16}{x^1}-64x^3 &= x^3&& \text{multiply ALL terms by } \color{blue}{ x^1 }. \\[1 em]x^1\cdot1+x^1\cdot8x^3+x^1\cdot\frac{16}{x^1}-x^1\cdot64x^3 &= x^1\cdot1x^3&& \text{cancel out the denominators} \\[1 em]x+8x^4+16-64x^4 &= x^4&& \text{simplify left side} \\[1 em]-56x^4+x+16 &= x^4&& \text{move all terms to the left hand side } \\[1 em]-56x^4+x+16-x^4 &= 0&& \text{simplify left side} \\[1 em]-57x^4+x+16 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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