Let $a$, $b$ be positive rational numbers such that $b$ is not the square of a rational number. Is there a way of showing that the only non-trivial subfield of ${\bf Q}(\sqrt[3]{a+\sqrt{b}})$ is ${\bf Q}(\sqrt{b})$ without computing the Galois group of the normal closure?

If there exists a subfield of degree 3, it must be ${\bf Q}(\sqrt[3]{a^2-b})$ by a norm argument, but I can't see how to conclude from this.