If $P=3^{2000}+3^{-2000}$ and $Q=3^{2000}-3^{-2000}$ then the value of $P^{2}-Q^{2}$ is
(A) $3^{4000}$
(B) $2 \times 3^{-4000}$
(C) 0
(D) $2 \times 3^{4000}$
(E) 4

Solution 1

$$
\begin{aligned}
P^{2}-Q^{2} & =(P+Q)(P-Q)=\left[\left(3^{2000}+3^{-2000}\right)+\left(3^{2000}-3^{-2000}\right)\right]\left[\left(3^{2000}+3^{-2000}\right)-\left(3^{2000}-3^{-2000}\right)\right] \\
& =\left(2.3^{2000}\right)\left(2.3^{-2000}\right) \\
& =4.3^{\circ} \\
& =4
\end{aligned}
$$

Solution 2

$$
P^{2}-Q^{2}=\left(3^{2000}+3^{-2000}\right)^{2}-\left(3^{2000}-3^{-2000}\right)^{2}
$$

$$
\begin{aligned}
& =3^{4000}+2.3^{2000} \cdot 3^{-2000}+3^{-4000}-3^{4000}+2.3^{2000} \cdot 3^{-2000}-3^{-4000} \\
& =3^{4000}+2.3^{\circ}+3^{-4000}-3^{4000}+2.3^{\circ}-3^{-4000} \\
& =4
\end{aligned}
$$

ANSWER: (E)
If $P=3^{2000}+3^{-2000}$ and $Q=3^{2000}-3^{-2000}$ then the value of $P^{2}-Q^{2}$ is
(A) $3^{4000}$
(B) $2 \times 3^{-4000}$
(C) 0
(D) $2 \times 3^{4000}$
(E) 4