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

$P+q=3,P^4+q^4=369$

$p^2+q^2=9-2pq$

$p^4+q^4+2{\left(pq\right)}^2=81+4{\left(pq\right)}^2-36pq$

$2{\left(pq\right)}^2-36\left(pq\right)-288=0$

$pq=24,pq=-6$

${\left(\frac{1}{p}+\frac{1}{q}\right)}^{-2}={\left(\frac{pq}{p+q}\right)}^2={\left(\frac{-6}{3}\right)}^2=2^2=4$