1. (Easy) Find the largest value of $x$ such that $\sqrt[3]{x}+\sqrt[3]{10-x}=1$.
Answer: $5+2 \sqrt{13}$
Solution: Cubing both sides of the given equation yields $x+3 \sqrt[3]{x(10-x)}(\sqrt[3]{x}+\sqrt[3]{10-x})+10-x=1$, which then becomes $10+3 \sqrt[3]{x(10-x)}=1$ or $\sqrt[3]{x(10-x)} \stackrel{(1)}{=}-3$. Cubing both sides of equation (1) gives $x(10-x)=-27$ or $x^{2}-10 x=27$. This means $(x-5)^{2}=52$ so $x=5 \pm 2 \sqrt{13}$ and the largest real solution is $x=5+2 \sqrt{13}$.