1. **State the problem:** Given that \(xy < 0\), \(\frac{1}{x^2} + \frac{1}{y^2} = 40\), and \(x + y = \frac{1}{3}\), find the value of \(\frac{1}{x^4} + \frac{1}{y^4}\). 2. **Express given terms in a useful form:** Start by introducing \(a = \frac{1}{x}\) and \(b = \frac{1}{y}\). Then \(\frac{1}{x^2} + \frac{1}{y^2} = a^2 + b^2 = 40\). 3. **Relate \(a\) and \(b\) using \(x + y = \frac{1}{3}\):** From \(x + y = \frac{1}{3}\), multiply both sides by \(ab = \frac{1}{xy}\): $$ (x + y)ab = \frac{1}{3} ab \implies (x + y) \left(\frac{1}{x} \cdot \frac{1}{y}\right) = \frac{1}{3} \cdot \frac{1}{xy} $$ But this is complicated; instead, note that: Since \(a = \frac{1}{x}\), \(b = \frac{1}{y}\), then \(x + y = \frac{1}{3} \implies \frac{1}{a} + \frac{1}{b} = \frac{1}{3}\). Multiply by \(ab\): $$ b + a = \frac{ab}{3} $$ 4. **Rewrite as equation in terms of \(a\) and \(b\):** \(a + b = \frac{ab}{3}\). 5. **Let \(S = a + b\) and \(P = ab\):** We have two pieces of information from the problem: - \(a^2 + b^2 = 40\), - \(a + b = S = \frac{P}{3}\). 6. **Relate \(a^2 + b^2\) to \(S\) and \(P\):** Since \(a^2 + b^2 = (a + b)^2 - 2ab = S^2 - 2P\), we get: $$ S^2 - 2P = 40 $$ But \(S = \frac{P}{3}\), so substitute: $$ \left(\frac{P}{3}\right)^2 - 2P = 40 \implies \frac{P^2}{9} - 2P = 40 $$ Multiply both sides by 9: $$ P^2 - 18P = 360 $$ Rearranged: $$ P^2 - 18P - 360 = 0 $$ 7. **Solve quadratic for \(P\):** $$ P = \frac{18 \pm \sqrt{18^2 + 4 \cdot 360}}{2} = \frac{18 \pm \sqrt{324 + 1440}}{2} = \frac{18 \pm \sqrt{1764}}{2} = \frac{18 \pm 42}{2} $$ Two solutions: - \(P = \frac{18 + 42}{2} = 30\) - \(P = \frac{18 - 42}{2} = -12\) Since \(xy < 0\), then \(ab = \frac{1}{xy}\) will have opposite sign. Here \(P = ab\) corresponds to \(\frac{1}{xy}\), so \(ab = P\). Opposite signs means \(P < 0\), so choose \(P = -12\). 8. **Find \(S\):** \(S = \frac{P}{3} = \frac{-12}{3} = -4\). 9. **Find desired value \(a^4 + b^4 = \frac{1}{x^4} + \frac{1}{y^4}\):** Recall: $$ a^4 + b^4 = (a^2 + b^2)^2 - 2a^2b^2 $$ From the problem: - \(a^2 + b^2 = 40\). Also, - \(a b = P = -12\), so \(a^2 b^2 = P^2 = (-12)^2 = 144\). Calculate: $$ a^4 + b^4 = 40^2 - 2 \times 144 = 1600 - 288 = 1312 $$ **Final answer:** \(\boxed{1312}\).