In this paper, The 1729 K3 Surface, the link is here https://arxiv.org/abs/1510.00735, Ken Ono writes
In particular, parametrized solutions to $X^3 + Y^3 = Z^3 + W^3$ give us families of elliptic curves with rank at least two. Ramanujan gave us such solutions in (1). He means the Ramanujan's identity $$(3a^2 + 5ab - 5b^2)^3 + (4a^2 - 4ab + 6b^2)^3 + (5a^2 - 5ab - 3b^2)^3 = (6a^2 - 4ab + 4b^2)^3$$
Let \begin{equation*} \begin{aligned} P1 &= (x_1(T ), y_1(T )) = (6T^2 - 4T + 4,\hspace{1mm} -3T^2 - 5T + 5)\\ P2 &= (x_2(T ), y_2(T )) = (4T^2 - 4T + 6, \hspace{2mm} 5T^2 - 5T - 3) \end{aligned} \end{equation*}
and $$k(T ) = 63(3T^2 - 3T + 1)(T^2 + T + 1)(T^2 - 3T + 3)$$
Then $$x_1(T)^3 + y_1(T)^3 = k(T ) = x_2(T)^3 + y_2(T)^3$$.
It is clear that Ramanujan's identity gives the solution. But what is k(T)? It looks discriminant. How this can be calculated in this case?