We have \[ p + q \mid (-q)^3 - q^3 = -2q^3 \ \text{and} \ p + q \mid 2p^3 \]Therefore, $p + q \mid 2 \gcd(p,q)^3$. If $p \not= q$, then $p + q \mid 2$, a contradiction -- and thus $p = q$.