$$2x^2+3x+\frac{8}{x^2}-\frac{6}{x}-13 = 0$$

$$ \begin{aligned} 2x^2+3x+\frac{8}{x^2}-\frac{6}{x}-13 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^2x }. \\[1 em]x^2x\cdot2x^2+x^2x\cdot3x+x^2x\cdot\frac{8}{x^2}-x^2x\cdot\frac{6}{x}-x^2x\cdot13 &= x^2x\cdot0&& \text{cancel out the denominators} \\[1 em]2x^5+3x^4+\frac{8}{x^1}-6-13x^3 &= 0&& \text{multiply ALL terms by } \color{blue}{ x^1 }. \\[1 em]x^1\cdot2x^5+x^1\cdot3x^4+x^1\cdot\frac{8}{x^1}-x^1\cdot6-x^1\cdot13x^3 &= x^1\cdot0&& \text{cancel out the denominators} \\[1 em]2x^6+3x^5+8-6x-13x^4 &= 0&& \text{simplify left side} \\[1 em]2x^6+3x^5-13x^4-6x+8 &= 0&& \\[1 em] \end{aligned} $$

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