History of Hindu Mathematics: A Source Book v2.c17.s24.11

v2.c17.s24.11

¹ Now m³ + n³ can be made a square by putting $m = (p³ + q³)p, n = (p³ + q³)q,$ so that $m^3 + n^3 = (p^3 + q^3)^4. Hence the solution of our equation will be x = (pr⁶)/((p² + q²)² (p³ + q³)³), y = (qr⁶)/((p² + q²)² (p³ + q³)³).$ Putting $r = (p^2 + q^2)(p^3 + q^3)s, we have the solution in positive integers as x = p(p² + q²)⁴ (p³ + q³)³ s⁶, y = q(p² + q²)⁴ (p³ + q³)³ s⁶;$ where $p$, $q$, $s$ are any integral numbers.