Given $$p + q = 3$$ and $$p^4 + q^4 = 369$$. Find $$\left(\frac{1}{p} + \frac{1}{q}\right)^{-2}$$.
First, we find $$pq$$. We know $$p + q = 3$$, so $$(p+q)^2 = 9 \implies p^2 + q^2 = 9 - 2pq$$. Now, $$(p^2 + q^2)^2 = p^4 + q^4 + 2p^2q^2$$, so $$(9 - 2pq)^2 = 369 + 2(pq)^2$$. Let $$t = pq$$: $$81 - 36t + 4t^2 = 369 + 2t^2$$, which simplifies to $$2t^2 - 36t + 81 - 369 = 0$$ and hence $$2t^2 - 36t - 288 = 0$$. Dividing by 2 gives $$t^2 - 18t - 144 = 0$$, so $$(t - 24)(t + 6) = 0$$. Since $$pq = 24$$ with $$p+q=3$$ leads to complex roots, we take $$pq = -6$$.
Next, we verify that this choice is correct. We have $$p^2 + q^2 = 9 - 2(-6) = 21$$, and therefore $$p^4 + q^4 = 21^2 - 2 \cdot 36 = 441 - 72 = 369$$, as required.
Finally, we compute the desired expression: $$\frac{1}{p} + \frac{1}{q} = \frac{p+q}{pq} = \frac{3}{-6} = -\frac{1}{2}$$, so $$\left(\frac{1}{p} + \frac{1}{q}\right)^{-2} = \left(-\frac{1}{2}\right)^{-2} = 4$$.
Hence the answer is $$4$$.
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