$\sqrt2$ and rational numbers

For $x + \sqrt{2}$ to be rational we will require that $x = a - \sqrt{2}$ for some rational number $a$ .

In this case $x^{2} + \sqrt{2} = (a - \sqrt{2})^{2} + \sqrt{2} = a^{2} - 2a\sqrt{2} + \sqrt{2} = a^{2} - (2a - 1)\sqrt{2}$ will be rational if $2a - 1 = 0 \Longrightarrow a = \dfrac{1}{2}$ .

So if $x = \dfrac{1}{2} - \sqrt{2}$ then statement A is correct.

Now with $x = a - \sqrt{2}$ we have $x^{3} + \sqrt{2} = (a - \sqrt{2})^{3} + \sqrt{2} = a^{3} - 3\sqrt{2}a^{2} + 6a - 2\sqrt{2} + \sqrt{2} = a^{3} + 6a - (3a^{2}+ 1)\sqrt{2}$ ,

which would be rational when $3a^{2} + 1 = 0$ . As this has no real solutions, we see that statement B cannot be satisfied, and so the correct answer is $\boxed{\text{Only A}}$ .