It is easy to find binary quadratic form parameterizations $F(x,y)$ to,

$$a^3+b^3+c^3+d^3 = 0\tag{1}$$

(See the identity (5) described in this MSE post.) To solve,

$$x_1^3+x_2^3+x_3^3 = 1\tag{2}$$

in the integers, all one has to do is to check if one term of $(1)$ can be solved as a *Pell-like* equation $F_i(x,y) = 1$. For example, starting with the cubes of the *taxicab number* 1728 as $a,b,c,d = 1,-9,12,-10$, we get,

$$a,b = x^2-11xy+9y^2,\;-9x^2+11xy-y^2$$

$$c,d = 12x^2-20xy+10y^2,\;-10x^2+20xy-12y^2$$

We can then solve $a = x^2-11xy+9y^2 = \pm 1$ since it can be transformed to the Pell equation $p^2-85q^2 = \pm 1$, thus giving an infinite number of integer solutions to $(2)$.

* Question*: How easy is it to find a quadratic form parameterization to,

$$x_1^3+x_2^3+x_3^3 = Nx_4^3\tag{3}$$

for $N$ a non-cube integer? I'm sure one can see where I'm getting at. If one can solve,

$$x_4 = c_1x^2+c_2xy+c_3y^2 = \pm 1\tag{4}$$

as a Pell-like equation, then that would prove that,

$$x_1^3+x_2^3+x_3^3 = N\tag{5}$$

is solvable in the integers in an *infinite* number of ways. (So far, this has only been shown for $N = 1,2$). The closest I've found is a *cubic* identity for $N = 3$ in a 2010 *paper* by Choudhry,

$$\begin{aligned} x_1 &= 2x^3+3x^2y+3xy^2\\ x_2 &= 3x^2y+3xy^2+2y^3\\ x_3 &= -2x^3-3x^2y-3xy^2-2y^3\\ x_4 &= xy(x+y) \end{aligned}$$

which is a special case of eq.58 in the paper.

Anybody knows how to find a quadratic form parametrization to $(3)$? (If one can be found, hopefully $(4)$ can also be solved.)