Independence of square roots

Assume. $a+b\sqrt2+c\sqrt3+d\sqrt6=0$.

Isolate. $d\sqrt6=-(a+b\sqrt2+c\sqrt3)$.

Square. $6d^2=a^2+2b^2+3c^2+2ab\sqrt2+2ac\sqrt3+2bc\sqrt6$.

Compare. $\sqrt6$ term gives $2bc=0$, i.e. $b=0$ or $c=0$. Say $b=0$. Then $a+c\sqrt3+d\sqrt6=0$, i.e. $a=-\sqrt3(c+d\sqrt2)$.

Square again. $a^2=3(c+d\sqrt2)^2=3(c^2+2d^2+2cd\sqrt2)$. Since $a,c,d\in\mathbb{Q}$ and $\sqrt2\notin\mathbb{Q}$, $cd=0$.

Case $c=0$: $a+d\sqrt6=0$ with $a,d\in\mathbb{Q}$, so $d=0$, $a=0$.

Case $d=0$: $a+c\sqrt3=0$, so $c=0$, $a=0$.

Conclusion. All coefficients are zero, so the four numbers are $\mathbb{Q}$-linearly independent.