Consider the equation: $$ \frac{1}{1+\sqrt[3]{2}}=a+b\sqrt[3]{2}+c\sqrt[3]{4} \quad a,b,c\in\mathbb{Q} $$ If $a,b,c$ were real, the equation would obviously have infinite solutions. But restricting them to the rationals leads to (supposedly) only one.
Using the identity
\begin{align} x^3-y^3&=(x-y)(x^2+xy+y^2)\quad x,y\in\mathbb{R}\\ \frac{x^3-y^3}{x-y}&=x^2+xy+y^2 \end{align} and choosing $x=1$ and $y=-\sqrt[3]{2}$, one can solve
\begin{align} \frac{3}{1+\sqrt[3]{2}}&=1-\sqrt[3]{2}+\sqrt[3]{4}\\ \frac{1}{1+\sqrt[3]{2}}&=\frac{1}{3}-\frac{1}{3}\sqrt[3]{2}+\frac{1}{3}\sqrt[3]{4} \end{align} it is then obvious that $a=\frac{1}{3}$, $b=-\frac{1}{3}$, $c=\frac{1}{3}$. The professor didn't explain further. My questions would be:
- Is there any clearer way to write this?
- Is there a universal method to solve this, other then "educated guessing" of $x$ and $y$?
- Is there any way to prove that this solution is indeed the only one?
Thank you.