Prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$ I'm trying to prove that $\sqrt{3} \notin \mathbb{Q}(\sqrt2)$
Suposse that $\sqrt{3}=a+b\sqrt{2}$
$\begin{align*}
\sqrt{3}&=a+b\sqrt{2}\\
3&=(a+b\sqrt{2})^2\\
3&=a^2+2\sqrt{2}ab+b^2\\
(3-a^2-12b^2)^2&=(2\sqrt{2}ab)^2\\
9-6a^2-12b^2+4a^2b^2+a^4+4b^4&=8a^2b^2
\end{align*}$
But I don't know what else I can do here.

I would start the same:
$$\sqrt3 = a + b\sqrt{2},\ \ a,b\in\mathbb{Q}$$
then square both to get
$$3 = a^2 + 2ab\sqrt2 + b^2$$
just like you and then rearrange to get
$$\frac{3 - a^2 - b^2}{2ab} = \sqrt{2}$$
meaning that $\sqrt2$ would have to be rational, which we know is not true.
