Three three three

The substitution $y^3 = x^3 t$ gives $\int_0^\infty \frac{1}{(x^3 + y^3)^3}\,dy \; = \; \frac{1}{3x^8}\int_0^\infty \frac{t^{-\frac23}}{(1 + t)^3}\,dt \; = \; \frac{1}{3x^8}B(\tfrac13,\tfrac83) \;,$ for any $x > 0$ , so the double integral is $\tfrac13B(\tfrac13,\tfrac83)\int_1^\infty x^{-8}\,dx \; = \; \tfrac{1}{21}B(\tfrac13,\tfrac83) \; = \; \frac{10\pi \sqrt{3}}{567} \;,$ making the answer $10 + 3 + 567 \,=\, \boxed{580}$ .