HongKong 2022-23 IMO Selection Tests

Answer: 4044 Let $a=u^3$ and $b=v^3$. Then we have $u^4-2022u^2+2023=0$ and $v^4-2022v^2+2023=0$, so $u^2$ and $v^2$ are the two roots of $t^2-2022t+2023=0$. In particular we have $u^2+v^2=2022$. Note also that $p+q=u^3+3u{v}^2+v^3+3u^{2}v=(u+v{)}^3$ and similarly $p-q=(u-v{)}^3$. It follows that $$(p+q{)}^{\frac{2}{3}}+(p-q{)}^{\frac{2}{3}}=(u+v{)}^2+(u-v{)}^2=2(u^2+v^2)=4044.$$