Find the complete integral of p3 +q^3 = 27z.
Let u=x+ayu= x+ayu=x+ay where a is an arbitrary constant. Replacing p by dzdu\frac{dz}{du}dudz and q by adzdua\frac{dz}{du}adudz
(dzdu)3+(adzdu)3=27z(dzdu)3+a3(dzdu)3=27z(1+a3)(dzdu)=27z1+a3zdz=27uIntegrating(1+a3)Log z=27u+b(1+a3)Log z=27(x+ay)+b(\frac{dz}{du})^3+(a\frac{dz}{du})^3= 27 z\\ (\frac{dz}{du})^3+a^3(\frac{dz}{du})^3= 27 z\\ (1+a^3)(\frac{dz}{du})=27z\\ \frac{1+a^3}{z}dz = 27u\\ Integrating \\ (1+a^3)Log \space z= 27 u +b\\ (1+a^3)Log \space z= 27 (x+ay) +b\\(dudz)3+(adudz)3=27z(dudz)3+a3(dudz)3=27z(1+a3)(dudz)=27zz1+a3dz=27uIntegrating(1+a3)Log z=27u+b(1+a3)Log z=27(x+ay)+b
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