Transform the fraction

$$
\frac{5 \sqrt[3]{6}-2 \sqrt[3]{12}}{4 \sqrt[3]{12}+2 \sqrt[3]{6}}
$$

so that the denominator is a rational number.
I. Solution: In our expression, $\sqrt[3]{6}$ can be factored out from every term in the numerator and the denominator:

$$
\frac{5 \sqrt[3]{6}-2 \sqrt[3]{12}}{4 \sqrt[3]{12}+2 \sqrt[3]{6}}=\frac{\sqrt[3]{6}(5-2 \sqrt[3]{2})}{\sqrt[3]{6}(4 \sqrt[3]{2}+2)}=\frac{5-2 \sqrt[3]{2}}{2(2 \sqrt[3]{2}+1)}
$$

The denominator will be rational if we multiply by $(4 \sqrt[3]{4}-2 \sqrt[3]{2}+1)$, because $(a+b)\left(a^{2}-a b+b^{2}\right)=a^{3}+b^{3}$:

$$
\begin{gathered}
\frac{5-2 \sqrt[3]{2}}{2(2 \sqrt[3]{2}+1)}=\frac{(5-2 \sqrt[3]{2})(4 \sqrt[3]{4}-2 \sqrt[3]{2}+1)}{2(2 \sqrt[3]{2}+1)(4 \sqrt[3]{4}-2 \sqrt[3]{2}+1)}= \\
=\frac{20 \sqrt[3]{4}-8 \cdot 2-10 \sqrt[3]{2}+4 \sqrt[3]{4}+5-2 \sqrt[3]{2}}{2(16+1)}=\frac{24 \sqrt[3]{4}-12 \sqrt[3]{2}-11}{34}
\end{gathered}
$$

With this, we have achieved our goal, we have "rationalized" the denominator.

Gábor Tusnády (Sátoraljaújhely, Kossuth g. II. o. t.)

II. Solution: Consider the simplified form of our expression by dividing by $\sqrt[3]{6}$, from which we have also omitted the $\frac{1}{2}$ factor: $\frac{5-2 \sqrt[3]{2}}{2 \sqrt[3]{2}+1}$

Let's try to bring this fraction to a form where the denominator is a single rational number, and the numerator contains at most the second power of $\sqrt[3]{2}$, that is, let's imagine the fraction in the following form, with the division of the coefficients already performed:

$$
\frac{5-2 \sqrt[3]{2}}{2 \sqrt[3]{2}+1}=a+b \sqrt[3]{2}+c \sqrt[3]{4}
$$

Multiplying through by the denominator:

$$
\begin{gathered}
5-2 \sqrt[3]{2}=2 a \sqrt[3]{2}+2 b \sqrt[3]{4}+4 c+a+b \sqrt[3]{2}+c \sqrt[3]{4}= \\
=(2 a+b) \sqrt[3]{2}+(2 b+c) \sqrt[3]{4}+(4 c+a)
\end{gathered}
$$

By comparing the coefficients on the right and left sides, it is clear that this only holds if

$$
\begin{aligned}
5 & =a+4 c \\
-2 & =2 a+b \\
0 & =2 b+c
\end{aligned}
$$

Solving the system of equations:

$$
a=-\frac{11}{17} \quad b=-\frac{12}{17} \quad c=\frac{24}{17}
$$

Substituting these back, we see that the rational denominator form of the fraction is the same as before.

Magda Bayer (Bp. XX., Bagi I. lg. II. o. t.)