SI. 2 If $\sqrt[3]{13 Q+6 P+1}-\sqrt[3]{13 Q-6 P-1}=\sqrt[3]{2}$, where $Q>0$, find the value of $Q$.
$$
\begin{array}{l}
(\sqrt[3]{13 Q+37}-\sqrt[3]{13 Q-37})^{3}=2 \\
13 Q+37-3 \sqrt[3]{(13 Q+37)^{2}} \sqrt[3]{13 Q-37}+3 \sqrt[3]{(13 Q-37)^{2}} \sqrt[3]{13 Q+37}-(13 Q-37)=2 \\
24=\sqrt[3]{(13 Q)^{2}-37^{2}} \sqrt[3]{13 Q+37}-\sqrt[3]{(13 Q)^{2}-37^{2}} \sqrt[3]{13 Q-37} \\
24=\sqrt[3]{(13 Q)^{2}-37^{2}} \sqrt[3]{2} ; \quad(\because \sqrt[3]{13 Q+37}-\sqrt[3]{13 Q-37}=\sqrt[3]{2}) \\
13824=\left[(13 Q)^{2}-1369\right] \times 2 \\
6912+1369=169 Q^{2} \\
Q^{2}=49 \Rightarrow Q=7 \\
\end{array}
$$

Method $2 \sqrt[3]{13 b+37}-\sqrt[3]{13 b-37}=\sqrt[3]{2}$,
We look for the difference of multiples of $\sqrt[3]{2}$
$\sqrt[3]{8 \times 2}-\sqrt[3]{2}=\sqrt[3]{2} \Rightarrow 13 b+37=16,13 b-37=2$, no solution
$\sqrt[3]{27 \times 2}-\sqrt[3]{8 \times 2}=\sqrt[3]{2} \Rightarrow 13 b+37=54,13 b-37=16$, no solution
$$
\sqrt[3]{64 \times 2}-\sqrt[3]{27 \times 2}=\sqrt[3]{2} \Rightarrow 13 b+37=128,13 b-37=54 \Rightarrow b=7
$$