Question
$$x-\frac{3}{6}x^2+\frac{1}{3}x^2 = \frac{1}{x^3}$$
Answer
This equation has no solution.
Explanation
$$ \begin{aligned} x-\frac{3}{6}x^2+\frac{1}{3}x^2 &= \frac{1}{x^3}&& \text{multiply ALL terms by } \color{blue}{ 6\cdot3x^3 }. \\[1 em]6\cdot3x^3x-6\cdot3x^3\frac{3}{6}x^2+6\cdot3x^3\frac{1}{3}x^2 &= 6\cdot3x^3\cdot\frac{1}{x^3}&& \text{cancel out the denominators} \\[1 em]18x^4-9x+6x &= 18&& \text{simplify left side} \\[1 em]18x^4-3x &= 18&& \text{move all terms to the left hand side } \\[1 em]18x^4-3x-18 &= 0&& \\[1 em] \end{aligned} $$
This polynomial has no rational roots that can be found using Rational Root Test.
Roots were found using quartic formulas
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