<s>[INST] How to rationalize denominator? Suppose $c$ is not a complete square integer, ${a_0},{a_1} \in \mathbb{Q}$, we have
$$
\frac{1}{{{a_0} + {a_1}\sqrt c }} 
= \frac{{{a_0} - {a_1}\sqrt c }}{{a_0^2 - a_1^2c}}.
$$
We need to show ${{a_0} + {a_1}\sqrt c } = 0$ iff ${a_0} = {a_1} = 0$. I know it's not hard.
Suppose $c$ is not a complete cube integer, ${a_0},{a_1},{a_2} \in \mathbb{Q}$, how can we deal with
$$
\frac{1}{{{a_0} + {a_1}\sqrt[3]{c} + {a_2}\sqrt[3]{{{c^2}}}}}
$$
similarly?
Any help will be appreciated.
 [/INST] I found a way. Let ${p^3} = c$, we write
$\left\{ \begin{gathered}
  {a_0} + {a_1}p + {a_2}{p^2} = S \hfill \\
  {a_2}c + {a_0}p + {a_1}{p^2} = pS \hfill \\
  {a_1}c + {a_2}cp + {a_0}{p^2} = {p^2}S \hfill \\ 
\end{gathered}  \right.$
So
$\left| {\begin{array}{*{20}{c}}
  {{a_0} - S}&{{a_1}}&{{a_2}} \\ 
  {{a_2}c - pS}&{{a_0}}&{{a_1}} \\ 
  {{a_1}c - {p^2}S}&{{a_2}c}&{{a_0}} 
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  {{a_0}}&{{a_1}}&{{a_2}} \\ 
  {{a_2}c}&{{a_0}}&{{a_1}} \\ 
  {{a_1}c}&{{a_2}c}&{{a_0}} 
\end{array}} \right| - S\left| {\begin{array}{*{20}{c}}
  1&{{a_1}}&{{a_2}} \\ 
  p&{{a_0}}&{{a_1}} \\ 
  {{p^2}}&{{a_2}c}&{{a_0}} 
\end{array}} \right| = 0$
And
$\frac{1}{S} = \left| {\begin{array}{*{20}{c}}
  1&{{a_1}}&{{a_2}} \\ 
  p&{{a_0}}&{{a_1}} \\ 
  {{p^2}}&{{a_2}c}&{{a_0}} 
\end{array}} \right|{\left| {\begin{array}{*{20}{c}}
  {{a_0}}&{{a_1}}&{{a_2}} \\ 
  {{a_2}c}&{{a_0}}&{{a_1}} \\ 
  {{a_1}c}&{{a_2}c}&{{a_0}} 
\end{array}} \right|^{ - 1}}$
But I cannot see why $\left| {\begin{array}{*{20}{c}}
  {{a_0}}&{{a_1}}&{{a_2}} \\ 
  {{a_2}c}&{{a_0}}&{{a_1}} \\ 
  {{a_1}c}&{{a_2}c}&{{a_0}} 
\end{array}} \right| \ne 0$.
</s>