$$x+\frac{1}{x}-1 = -\frac{9}{x}+3+\frac{8}{x^2}+2x-3$$

$$ \begin{aligned} x+\frac{1}{x}-1 &= -\frac{9}{x}+3+\frac{8}{x^2}+2x-3&& \text{multiply ALL terms by } \color{blue}{ xx^2 }. \\[1 em]xx^2x+xx^2\cdot\frac{1}{x}-xx^2\cdot1 &= -xx^2\cdot\frac{9}{x}+xx^2\cdot3+xx^2\cdot\frac{8}{x^2}+xx^2\cdot2x-xx^2\cdot3&& \text{cancel out the denominators} \\[1 em]x^4+1-x^3 &= -9+3x^3+\frac{8}{x^1}+2x^4-3x^3&& \text{multiply ALL terms by } \color{blue}{ x^1 }. \\[1 em]x^1\cdot1x^4+x^1\cdot1-x^1\cdot1x^3 &= -x^1\cdot9+x^1\cdot3x^3+x^1\cdot\frac{8}{x^1}+x^1\cdot2x^4-x^1\cdot3x^3&& \text{cancel out the denominators} \\[1 em]x^5+x-x^4 &= -9x+3x^4+8+2x^5-3x^4&& \text{simplify left and right hand side} \\[1 em]x^5-x^4+x &= -9x+3x^4+8+2x^5-3x^4&& \\[1 em]x^5-x^4+x &= 2x^5-9x+8&& \text{move all terms to the left hand side } \\[1 em]x^5-x^4+x-2x^5+9x-8 &= 0&& \text{simplify left side} \\[1 em]-x^5-x^4+10x-8 &= 0&& \\[1 em] \end{aligned} $$

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