1. Find the integer part of $(\sqrt{3}+1)^{6}$.

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1. Solution Let $A=(\sqrt{3}+1)^{6}, B=(\sqrt{3}-1)^{6}$, consider
$$
A+B=(\sqrt{3}+1)^{6}+(\sqrt{3}-1)^{6}=2\left[(\sqrt{3})^{6}+C_{6}^{2}(\sqrt{3})^{4}+C_{6}^{4}(\sqrt{3})^{2}+1\right]=416
$$

Since $0<\sqrt{3}-1<1$, we have $0<(\sqrt{3}-1)^{6}<1$. Therefore, the integer part of $(\sqrt{3}+1)^{6}$ is 415.