# 2010 Australian Mathematical Olympiad, Problem 6

Solution by Steve Dinh, a.k.a. Vo Duc Dien

Prove that

\sqrt[3]{{6}+\sqrt[3]{845}+\sqrt[3]{325}}+\sqrt[3]{{6}+\sqrt[3]{847}+\sqrt[3]{539}} = \sqrt[3]{{4}+\sqrt[3]{245}+\sqrt[3]{175}}+\sqrt[3]{{8}+\sqrt[3]{1859}+\sqrt[3]{1573}}.

Observe that

845 = 13²·5
325 = 5²·13
847 = 11²·7
539 = 7²·11
245 = 7²·5
175 = 5²·7
1859 = 13²·11
1573 = 11²·13

and

6+\sqrt[3]{845}+\sqrt[3]{325} = \bigg(\sqrt[3]{\frac{13}{3}}\bigg)^3 + 3\bigg(\sqrt[3]{\frac{13}{3}}\bigg)^{2}\sqrt[3]{\frac{5}{3}}+ 3\sqrt[3]{\frac{13}{3}}\bigg(\sqrt[3]{\frac{5}{3}}\bigg)^{2} + \bigg(\sqrt[3]{\frac{5}{3}}\bigg)^{3},

which is to say

\sqrt[3]{6+\sqrt[3]{845}+\sqrt[3]{325}} = \sqrt[3]{\frac{13}{3}} + \sqrt[3]{\frac{5}{3}}.

Similarly,

\sqrt[3]{6+\sqrt[3]{847}+\sqrt[3]{539}} = \sqrt[3]{\frac{11}{3}} + \sqrt[3]{\frac{7}{3}}.
\sqrt[3]{4+\sqrt[3]{245}+\sqrt[3]{175}} = \sqrt[3]{\frac{7}{3}} + \sqrt[3]{\frac{5}{3}}.
\sqrt[3]{8+\sqrt[3]{1859}+\sqrt[3]{1573}} = \sqrt[3]{\frac{13}{3}} + \sqrt[3]{\frac{11}{3}}.

Comparing the right hand sides, we obtain the required identity.