smc

Let $a=\sqrt[3]{x+9}$ and $b=\sqrt[3]{x-9}$. Cubing these equations, we get $a^3=x+9$ and $b^3=x-9$, so $a^3-b^3=18$. The left-hand side factors as $$(a-b)(a^2+ab+b^2)=18.$$ However, from the given equation $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$, we get $a-b=3$. Then $3(a^2+ab+b^2)=18$, so $a^2+ab+b^2=18/3=6$. Squaring the equation $a-b=3$, we get $a^2-2ab+b^2=9$. Subtracting this equation from the equation $a^2+ab+b^2=6$, we get $3ab=-3$, so $ab=-1$. But $a=\sqrt[3]{x+9}$ and $b=\sqrt[3]{x-9}$, so $ab=\sqrt[3]{(x+9)(x-9)}=\sqrt[3]{x^2-81}$, so $\sqrt[3]{x^2-81}=-1$. Cubing both sides, we get $x^2-81=-1$, so $x^2=80$. The answer is $\boxed{75\text{ and }85}$.