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Question
$$\frac{1}{x^2}-\frac{1}{x^3}+x^2 = \frac{3}{x}+1$$
Answer
This equation has no solution.
Explanation
$$ \begin{aligned} \frac{1}{x^2}-\frac{1}{x^3}+x^2 &= \frac{3}{x}+1&& \text{multiply ALL terms by } \color{blue}{ x^2x^3x }. \\[1 em]x^2x^3x\cdot\frac{1}{x^2}-x^2x^3x\cdot\frac{1}{x^3}+x^2x^3xx^2 &= x^2x^3x\cdot\frac{3}{x}+x^2x^3x\cdot1&& \text{cancel out the denominators} \\[1 em]\frac{1}{x^2}-\frac{1}{x^3}+x^8 &= 3x+x^6&& \text{multiply ALL terms by } \color{blue}{ x^2x^3 }. \\[1 em]x^2x^3\cdot\frac{1}{x^2}-x^2x^3\cdot\frac{1}{x^3}+x^2x^3\cdot1x^8 &= x^2x^3\cdot3x+x^2x^3\cdot1x^6&& \text{cancel out the denominators} \\[1 em]\frac{1}{x^1}-\frac{1}{x^2}+x^{13} &= 3x^6+x^{11}&& \text{multiply ALL terms by } \color{blue}{ x^1x^2 }. \\[1 em]x^1x^2\cdot\frac{1}{x^1}-x^1x^2\cdot\frac{1}{x^2}+x^1x^2\cdot1x^{13} &= x^1x^2\cdot3x^6+x^1x^2\cdot1x^{11}&& \text{cancel out the denominators} \\[1 em]1-\frac{1}{x^1}+x^{16} &= 3x^9+x^{14}&& \text{multiply ALL terms by } \color{blue}{ x^1 }. \\[1 em]x^1\cdot1-x^1\cdot\frac{1}{x^1}+x^1\cdot1x^{16} &= x^1\cdot3x^9+x^1\cdot1x^{14}&& \text{cancel out the denominators} \\[1 em]x-1+x^{17} &= 3x^{10}+x^{15}&& \text{simplify left and right hand side} \\[1 em]x^{17}+x-1 &= x^{15}+3x^{10}&& \text{move all terms to the left hand side } \\[1 em]x^{17}+x-1-x^{15}-3x^{10} &= 0&& \text{simplify left side} \\[1 em]x^{17}-x^{15}-3x^{10}+x-1 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using Newton method.
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