Express the $x^{12}-3^{6} y^{12}$ expression as the sum of cubes of two two-term expressions with integer coefficients.
Both terms of our expression are "perfect cubes" (rational integer expressions with integer coefficients raised to the third power): $x^{12}=\left(x^{4}\right)^{3},-3^{6} y^{12}=$ $\left(-3^{2} y^{4}\right)^{3}$. Therefore, we can try to find a solution where each term of the two binomial expressions is $x^{4}$ and $-3^{2} y^{4}$, respectively, or in other words, a (non-zero) $Z$ and $V$ monomial, such that:

$$
\left(x^{4}+Z\right)^{3}+\left(V-3^{2} y^{4}\right)^{3} \equiv x^{12}-3^{6} y^{4}
$$

or, expanding:

$$
\left(3 x^{8} Z+3 x^{4} Z^{2}+Z^{3}\right)+\left(V^{3}-3^{3} V^{2} y^{4}+3^{5} V y^{8}\right) \equiv 0
$$

Let's try with $Z$ and $V$ that contain only positive integer powers of $x$ and $y$. Then the first trinomial is certainly arranged in ascending powers of $y$, and the terms in the second parenthesis also contain either ascending or descending powers of $y$, because the 2nd and 3rd terms' variable part is $y^{4} / V$ times the one preceding it. Thus, for the identity to hold, the two middle terms must be negatives of each other:

$$
3 x^{4} Z^{2}=3^{3} V^{2} y^{4}, \quad \text { or } \quad Z^{2} / V^{2}=3^{2} y^{4} / x^{4} \text {. }
$$

Accordingly, it is impossible for the terms in the second parenthesis to be arranged in descending powers of $y$, because this assumption, through $Z^{3}=-V^{3}$, leads to $Z^{3} / V^{3}=-1$ and $Z^{2} / V^{2}=1$, so (2) can only be satisfied for certain value pairs of $x$ and $y$, and thus (1) cannot hold for any value pair of $x$ and $y$. The other possibility, the

$$
3 x^{8} Z=-V^{3} \quad \text { and } \quad \text { (4) } \quad Z^{3}=-3^{5} V y^{8}
$$

equalities do not lead to a contradiction with (2), because multiplying the corresponding sides of (4) yields $Z^{4} / V^{4}=3^{4} y^{8} / x^{8}$, and this is a consequence of (2). Therefore, (4) can be omitted.

Now, substituting the expression for $Z$ obtained from (3) into (2), we get: $V= \pm 3 x^{3} y$, and thus $Z=\mp 3^{2} x y^{3}$.

Both solutions meet our requirements, as

$$
x^{12}-3^{6} y^{12} \equiv\left(x^{4}-3^{2} x y^{3}\right)^{3}+\left(3 x^{3} y-3^{2} y^{4}\right)^{3} \equiv\left(x^{4}+3^{2} x y^{3}\right)^{3}+\left(-3 x^{3} y-3^{2} y^{4}\right)^{3}
$$

Ferenc Mezey (Budapest, Rákóczi F. g. III. o. t)