Ukrainian National Mathematical Olympiad

We rewrite the equation as follows.
$$n^3 = p^3 + 2p^2q + 2pq^2 + q^3 = (p+q)(p^2 + pq + q^2)$$
Clearly, $p \neq q$, because, otherwise, the equation $n^3 = 6p^3$ would hold, which is not possible for positive integers.
Suppose there exists such $s > 1$, which is a factor of both $p+q$ and $(p^2 + pq + q^2)$. But then $s \mid (p+q)^2 - (p^2 + pq + q^2) = pq$. If, for example, $s \mid p$, then from $s \mid (p+q)$ follows $s \mid q \Rightarrow$ from $p, q$ being prime, it must be that $p = q$, which leads to contradiction.
If, on the other hand, $((p+q), (p^2 + pq + q^2)) = 1$, then each factor must be a cube of a positive integer. Suppose $p+q = x^3$, $p^2 + pq + q^2 = y^3$, and $n = xy$. Then,
$$pq = (p+q)^2 - (p^2 + pq + q^2) = x^6 - y^3 = (x^2 - y)(x^4 + x^2y + y^2).$$
Since $x^4 + x^2y + y^2 > x^3 = p+q$, we obtain $x^4 + x^2y + y^2 = pq$ and $x^2 - y = 1 \Rightarrow pq = x^4 + x^2(x^2-1) + (x^2-1)^2 = 3x^4 - 3x^2 + 1 = 3x^2(x-1)(x+1) + 1 \equiv 1 \pmod 9$, since $x(x-1)(x+1) \not\equiv 3 \pmod 9$.
What is left is to search through the cases modulo 9. Cube of an integer modulo 9 can be equal to 0, $\pm$ 1. Let us list all possible cases for remainders modulo 9 of primes $p, q$ so that condition $pq \equiv 1 \pmod 9$ is satisfied.
$$p \equiv 1 \pmod 9 \Rightarrow q \equiv 1 \pmod 9 \Rightarrow x^3 = p+q \equiv 2 \pmod 9 \{\text{ -- contradiction.}\}$$
$$p \equiv 2 \pmod 9 \Rightarrow q \equiv 5 \pmod 9 \Rightarrow x^3 = p+q \equiv 7 \pmod 9 \{\text{ -- contradiction.}\}$$
$$p \equiv 4 \pmod 9 \Rightarrow q \equiv 7 \pmod 9 \Rightarrow x^3 = p+q \equiv 2 \pmod 9 \{\text{ -- contradiction.}\}$$
$$p \equiv 8 \pmod 9 \Rightarrow q \equiv 8 \pmod 9 \Rightarrow x^3 = p+q \equiv 7 \pmod 9 \{\text{ -- contradiction.}\}$$
All the cases were checked, which concludes the proof that such numbers do not exist.