Finding All u in R for Field Extension Q(√2,³ √5) and Proving Equality

[cantgetit]

how do I find all u in R such that Q(u)=Q(√2,³ √5 ) (square root of two and cubed root of 5) and prove they are the same. PLEASE help I am desperate! I understand that one possible u is √2+³ √5 but i don't know hoe to prove it withou pages of algebra and i don't know what the others are.

 

[VKint]

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
$$u = a \alpha + b \beta + c \alpha \beta + d \beta^2 + e \alpha \beta^2$$
(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 $$x_1 + x_2$$?)