<s>[INST] On integers as sums of three integer cubes revisited 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.)
 [/INST] Perhaps if you start with my three-rational-cubes identity
$$
  ab^2 = \biggl(\frac{(a^2+3b^2)^3+(a^2-3b^2)(6ab)^2}{6a(a^2+3b^2)^2}\biggr)^{\!3}
    - \biggl(\frac{(a^2+3b^2)^2-(6ab)^2}{6a(a^2+3b^2)}\biggr)^{\!3}
    - \biggl(\frac{(a^2-3b^2)6ab^2}{(a^2+3b^2)^2}\biggr)^{\!3}
$$
you might be able to find something?
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