$$7x^4-4x^2+5-\frac{3}{x^3} = 0$$
Answer
$$ \begin{matrix}x_1 = 0.81881 & x_2 = 0.73123+0.66035i & x_3 = 0.73123-0.66035i \\[1 em] x_4 = -0.26307+0.68395i & x_5 = -0.26307-0.68395i & x_6 = -0.87756+0.4837i \\[1 em] x_7 = -0.87756-0.4837i & \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} 7x^4-4x^2+5-\frac{3}{x^3} &= 0&& \text{multiply ALL terms by } \color{blue}{ x^3 }. \\[1 em]x^3\cdot7x^4-x^3\cdot4x^2+x^3\cdot5-x^3\cdot\frac{3}{x^3} &= x^3\cdot0&& \text{cancel out the denominators} \\[1 em]7x^7-4x^5+5x^3-3 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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