Question 563151
You can let *[tex \LARGE \sqrt{a} + \sqrt[3]{b} + \sqrt[3]{c} = k] where k is an integer, then *[tex \LARGE \sqrt[3]{b} + \sqrt[3]{c} = k - \sqrt{a}]. Cubing both sides,


*[tex \LARGE b + c + 3\sqrt[3]{b^2c} + 3\sqrt[3]{bc^2} = k^3 - 3k^2\sqrt{a} + 3ka - a\sqrt{a}]


Here, we can take this equation "modulo 1" by eliminating all the integer expressions (b,c,k^3,3ka).


*[tex \LARGE 3\sqrt[3]{b^2c} + 3\sqrt[3]{bc^2} \equiv -3k^2\sqrt{a} - a\sqrt{a} (mod 1)]


Factor LHS


*[tex \LARGE 3\sqrt[3]{bc}(\sqrt[3]{b} + \sqrt[3]{c}) \equiv -3k^2\sqrt{a} - a\sqrt{a} (mod 1)] Note that you can replace *[tex \LARGE \sqrt[3]{b} + \sqrt[3]{c}] with *[tex \LARGE k - \sqrt{a}]. Modulo 1, this is equivalent to -sqrt(a).


*[tex \LARGE 3\sqrt[3]{bc}(-\sqrt{a}) \equiv -3k^2\sqrt{a} - a\sqrt{a} (mod 1)]


*[tex \LARGE 3\sqrt[3]{bc} \equiv a - 3k^2 \equiv 0 (mod 1)]


Here, we show that *[tex \LARGE 3\sqrt[3]{bc} \equiv 0 (mod 1)] i.e. it is an integer. I'll let you finish the proof that sqrt{a}, sqrt[3]{b}, and sqrt[3]{c} have to be integers. Pretty daunting problem...unfortunately we cannot assume the converse of the statement is true (i.e. if ... are integers then ... is an integer).