13. Find the largest real number $x$ such that $\sqrt[3]{x}+\sqrt[3]{4-x}=1$.
Answer: $2+\sqrt{5}$
Solution: Cubing both sides of the given equation yields $x+3 \sqrt[3]{x(4-x)}(\sqrt[3]{x}+\sqrt[3]{4-x})+4-x=1$, which then becomes $4+3 \sqrt[3]{x(4-x)}=1$ or $\sqrt[3]{x(4-x)} \stackrel{(1)}{=}-1$. Cubing both sides of equation (1) gives $x(4-x)=-1$ or $x^{2}-4 x=1$. This means $(x-2)^{2}=5$ so $x=2 \pm \sqrt{5}$ and the largest real solution is $x=2+\sqrt{5}$.