Using this intriguing equation, Ramanujan found the identity,
(3x^2+5xy-5y^2)^3+(4x^2-4xy+6y^2)^3+(5x^2-5xy-3y^2)^3 = (6x^2-4xy+4y^2)^3
and asked for more examples of solving cubic quadruples a^3+b^3+c^3+d^3 = 0 in terms of quadratic forms.
We answer this question by providing a nice general identity that, starting with any quadruple (a,b,c,d), can generate a quadratic form identity. Furthermore, it can go beyond quadruples so we can use other interesting equations like
11^3+12^3+13^3+14^3 = 20^3, and,
31^3+33^3+35^3+37^3+39^3+41^3 = 66^3.
(New!): "A Collection of Algebraic Identities" by this author.
