As number fields, we have that $\mathbb{Q}(\sqrt{2}+ \sqrt{3})\simeq \mathbb{Q}(\sqrt{2}, \sqrt{3}) $ however we can write the elements different says:
\begin{eqnarray*} x &=& a + b(\sqrt{2}+\sqrt{3}) + c(\sqrt{2}+\sqrt{3})^ 2 + d(\sqrt{2}+\sqrt{3})^3\\ &\text{or}& \\ x &=& a + b \sqrt{2} + c \sqrt{3} + d \sqrt{6} \end{eqnarray*}
It's not instantly obvious to me that $\sqrt{2} - \sqrt{3} \in \mathbb{Q}(\sqrt{2}+\sqrt{3})$ but I'm sure it' s the case...
What is the change of basis between one and the other? Maybe that way we can find the linear combination... I' d like the change of basis in order toshow:
$$ \sqrt{2}-\sqrt{3} = a + b\big(\sqrt{2}+\sqrt{3}\big) + c\big(\sqrt{2}+\sqrt{3}\big)^ 2 + d\big(\sqrt{2}+\sqrt{3}\big)^3 $$
These scratchworks easily turn into an answer:
\begin{eqnarray*} 1 &=& 1\\ \sqrt{2} + \sqrt{3} &=& 1 \cdot \sqrt{2} + 1 \cdot \sqrt{3} \\ (\sqrt{2} + \sqrt{3})^2 &=& 5 \cdot 1 + 2 \cdot \sqrt{6} \\ (\sqrt{2} + \sqrt{3})^3 &=& 11 \cdot \sqrt{2} + 5 \cdot \sqrt{3} \end{eqnarray*}
and we can get a change of basis matrix finally (as a matrix over $\mathbb{Q})$:
$$\left[ \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 5 & 0 & 0 & 2 \\ 0 & 11 & 5 & 0 \end{array}\right]\; \left[\begin{array}{c} 1 \\ \sqrt{2} \\ \sqrt{3} \\ \sqrt{6}\end{array}\right] = \left[\begin{array}{c} 1 \\ \;\sqrt{2} + \sqrt{3}\;\\ (\sqrt{2} + \sqrt{3})^2 \\ (\sqrt{2} + \sqrt{3})^3 \end{array} \right]$$
Incidentally I noticed that $\cos 15^\circ = \frac{\sqrt{6}- \sqrt{2}}{4} \in \mathbb{Q}(\sqrt{2}, \sqrt{3}) \simeq \mathbb{Q}(\sqrt{2} + \sqrt{3})$ using the difference of angles formula.