Let p and q be two real numbers such that p + q =3 and $p^4+q^4=369$ Then ${\left(\frac{1}{p}+\frac{1}{q}\right)}^{-2}$ is equal to ___________.

$$\begin{gathered} given that p+q=3 ,p^4+q^4=369; \\ (p+q{)}^2=9 \\ {p}^2+q^2=9-2pq \\ \frac{1}{{\left(\frac{1}{p}+\frac{1}{q}\right)}^2}=\frac{(qp{)}^2}{(q+p{)}^2}=\frac{(qp{)}^2}{9}; \\ now p^4+q^4={\left(p^2+q^2\right)}^2-2p^2{q}^2 \\ \Rightarrow 369=(9-2pq{)}^2-2(pq{)}^2 \\ \Rightarrow 369=81+4p^2{q}^2-36pq-2p^2{q}^2 \\ \Rightarrow 288=2p^2{q}^2-36pq \\ \Rightarrow 144=p^2{q}^2-18pq \\ (pq{)}^2-2\times 9\times pq+9^2=144+9^2 \\ (pq-9{)}^2=225 \\ pq-9=\pm 15 \\ pq=\pm 15+9 \\ pq =24,-6 \end{gathered}$$

(24 is rejected because $p^2+q^2=9-2pq$ is negative)

$\frac{(qp{)}^2}{9}=\frac{16\times 16}{9}=4$