I5.1 If $\left(\frac{1 \times 2 \times 4+2 \times 4 \times 8+3 \times 6 \times 12+\cdots+1999 \times 3998 \times 7996}{1^{3}+2^{3}+3^{3}+\cdots+1999^{3}}\right)^{\frac{1}{3}}=P$, find the value of $P$.
I5.2 If $(x-P)(x-2 Q)-1=0$ has two integral roots, find the value of $Q$.
$$
\begin{array}{l} 
P=\left(\frac{1 \times 2 \times 4+2 \times 4 \times 8+3 \times 6 \times 12+\cdots+1999 \times 3998 \times 7996}{1^{3}+2^{3}+3^{3}+\cdots+1999^{3}}\right)^{\frac{1}{3}} \\
=\left[\frac{1 \times 2 \times 4\left(1^{3}+2^{3}+3^{3}+\cdots+1999^{3}\right)}{1^{3}+2^{3}+3^{3}+\cdots+1999^{3}}\right]^{\frac{1}{3}} \\
=8^{\frac{1}{3}}=2 \\
(x-2)(x-2 Q)-1=0 \\
x^{2}-2(1+Q) x+4 Q-1=0
\end{array}
$$

Two integral roots $\Rightarrow \Delta$ is perfect square
$$
\begin{aligned}
\Delta & =4\left[(1+Q)^{2}-(4 Q-1)\right] \\
& =4\left(Q^{2}-2 Q+2\right) \\
& =4(Q-1)^{2}+4
\end{aligned}
$$

It is a perfect square $\Rightarrow Q-1=0, Q=1$
Method 2
$$
\begin{array}{l}
(x-2)(x-2 Q)=1 \\
(x-2=1 \text { and } x-2 Q=1) \text { or }(x-2=-1 \text { and } x-2 Q=-1) \\
(x=3 \text { and } Q=1) \text { or }(x=1 \text { and } Q=1) \\
\therefore Q=1
\end{array}
$$