XVI OBM

$a^3=a+1>1$, so $a>1$, so $b^6=b+3a>3$, so $b>1$. $a^6-b^6=(a+1{)}^2-(b+3a)=(a-1{)}^2+(a-b)>a-b$. But $a^6-b^6=(a-b)(a^5+a^{4}b+a^3{b}^2+a^2{b}^3+a{b}^4+b^5)$, so if $b>a$, then $b^6-a^6\ge 6(b-a)$ and hence $a^6-b^6<a-b$. Contradiction. Obviously if $a=b$, then $a^6=b^6$ and so $a^6-b^6$ is not greater than $a-b$. Hence we must have $a>b$.