Question
$$x^4-x^5-x+\frac{1}{81} = 0$$
Answer
$$ x = \dfrac{ 1 }{ 81 } $$
Explanation
$$ \begin{aligned} x^4-x^5-x+\frac{1}{81} &= 0&& \text{multiply ALL terms by } \color{blue}{ 81 }. \\[1 em]81x^4-81x^5-81x+81\cdot\frac{1}{81} &= 81\cdot0&& \text{cancel out the denominators} \\[1 em]81x^4-81x^5-81x+1 &= 0&& \text{simplify left side} \\[1 em]-81x^5+81x^4-81x+1 &= 0&& \\[1 em] \end{aligned} $$
This page was created using
Polynomial Equations Solver