Well, the degree of \mathbb{Q}(\sqrt{2}, \sqrt[3]{5}) over \mathbb{Q} is 6. (Prove this.) Thus, we know that if \{ 1, \alpha \} and \{ 1, \beta, \beta^2 \} are bases for \mathbb{Q}(\sqrt{2}) and \mathbb{Q}(\sqrt[3]{5}), respectively, then \{ 1, \alpha, \beta, \alpha \beta, \beta^2, \alpha \beta^2 \} is a basis for \mathbb{Q}(\sqrt{2}, \sqrt[3]{5}) over \mathbb{Q}. In particular, we can write
<br />
u = a \alpha + b \beta + c \alpha \beta + d \beta^2 + e \alpha \beta^2<br />
(why can we assume that the 1-component of u is 0 WLOG?). At this point, since you know the degrees of all of the basis elements over \mathbb{Q}, you should be able to find conditions on the coefficients a,b,c,d,e that guarantee that [ \mathbb{Q}(u) : \mathbb{Q} ] = 6, which is enough to show that \mathbb{Q}(u) = \mathbb{Q}(\sqrt{2}, \sqrt[3]{5}). (HINT: If the degrees of x_1 and x_2 over \mathbb{Q} are d_1, d_2, and \gcd(d_1, d_2) = 1, what is the degree of x_1 + x_2?)
