Example 7. Find the integer part of $(\sqrt{3}+1)^{6}$.

Translate the above text into English, please keep the original text's line breaks and format, and output the translation result directly.
$$
\begin{array}{l}
\text{Solve for the dual number } B = (\sqrt{3}-1)^{6} \text{ of the number } A = (\sqrt{3}+1)^{6}, \text{ and examine} \\
A+B=(\sqrt{3}+1)^{6}+(\sqrt{3}-1)^{6} \\
= \left[(\sqrt{3}+1)^{3}\right]^{2}+\left[(\sqrt{3}-1)^{3}\right]^{2} \\
= \left[(\sqrt{3}+1)^{3}-(\sqrt{3}-1)^{3}\right] \\
+2[(\sqrt{3}+1)(\sqrt{3}-1)]^{3} \\
=\{[(\sqrt{3}+1)-(\sqrt{3}-1)][(\sqrt{3} \\
+1)^{2}+(\sqrt{3}+1)(\sqrt{3}-1) \\
\left.\left. +(\sqrt{3}-1)^{2}\right]\right\}+16 \\
= 2^{2}(4+2 \sqrt{3}+2+4-2 \sqrt{3})^{2}+16 \\
= 416 . \\
\because 0<\sqrt{3}-1<1, \\
\therefore 0<(\sqrt{3}-1)^{6}<1,
\end{array}
$$

Thus, the integer part of $(\sqrt{3}+1)^{6}$ is 415. Note, the dual form of $(a+b)^{n}$ is $(a-b)^{n}$.