Without using a calculator or a table, determine the smallest integer greater than $(\sqrt{3}+\sqrt{2})^{6}$.
# Solution.

$$
(\sqrt{3}+\sqrt{2})^{6}=\left[(\sqrt{3}+\sqrt{2})^{3}\right]^{2}=485+198 \sqrt{6}
$$

If we were to determine the value of $(\sqrt{3}-\sqrt{2})^{6}$, the result would be $485-198 \sqrt{6}$. Since $\sqrt{3}$ and $\sqrt{2}$ are both less than $\sqrt{4}$ and greater than $\sqrt{1}$, they fall between 1 and 2:

$$
0<\sqrt{3}-\sqrt{2}<1
$$

This is also true for $(\sqrt{3}-\sqrt{2})^{6}$:

$$
0<(\sqrt{3}-\sqrt{2})^{6}<1
$$

or

$$
0<485-198 \sqrt{6}<1 .
$$

From this, it follows that

$$
484<198 \sqrt{6}<485
$$

Therefore, the smallest integer greater than $(\sqrt{3}+\sqrt{2})^{6}$ is

$$
485+485=970 .
$$