Does the 3.3.3 equation $x_1^3+x_2^3+x_3^3=y_1^3+y_2^3+y_3^3$ have a complete integer or rational solution? Consider the equation
$$
X_1^3+X_2^3+X_3^3 = Y_1^3+Y_2^3+Y_3^3.  \tag{$\star$}
$$
Does this have a complete solution (a.k.a. parameterization) in integers? If not, does it have a complete solution in rational numbers?
I know the [integral and/or rational] solutions to other related equations — such as the 3.2.2 equation
$$
X_1^3+X_2^3 = Y_1^3+Y_2^3
$$
and the 3.1.3 equation
$$
X_1^3 = Y_1^3+Y_2^3+Y_3^3.
$$
(see, for example, Choudhry’s paper) — but I have yet to find any paper giving the complete rational or integral solution to ($\star$).


Update:

Yes, there is a complete rational solution. Since we are dealing with a homogeneous equation, it is best to consider rational points on,
$$x_1^2+x_2^2 =1\tag1$$
$$x_1^3+x_2^3+x_3^3 =1\tag2$$
$$x_1^3+x_2^3+x_3^3+x_4^3+x_5^3 =1\tag3$$
to reduce the number of variables. The complete rational solution to $(1)$ and $(2)$ are well-known. It turns out the complete rational solution for $(3)$ is,
$$\color{blue}{\big(a + (a + b)\gamma\big)^3 + (a \gamma - b)^3 + \big(c + (c + d)\gamma\big)^3 + (c \gamma - d)^3 + \gamma^3 = 1}\tag4$$
where,
$$\gamma=\frac{-a^3 + b^3 - c^3 + d^3 + 1}{a^3 + (a+b)^3+c^3+(c+d)^3+1}\tag5$$
and can be found by generalizing the approach used by Choudhry in the paper cited by the OP.

Proof: 

Equating the five terms of $(3)$ and $(4)$, one can easily solve for the five unknowns $a,b,c,d,\gamma$ as,
$$a = \frac{x_1+x_2x_5}{\beta},\quad b = \frac{-x_2+(x_1-x_2)x_5}{\beta}$$
$$c = \frac{x_3+x_4x_5}{\beta},\quad d = \frac{-x_4+(x_3-x_4)x_5}{\beta}$$
$$\gamma = x_5, \quad\quad\beta = x_5^2+x_5+1$$
Substituting these into $(5)$, the equality holds only if the $x_i$ satisfies $x_1^3+x_2^3+x_3^3+x_4^3+x_5^3 =1$, thus proving the assertion.

Example:

The smallest solution to $(3)$ is,
$$\big(\tfrac{1}{4})^3+\big(\tfrac{1}{4})^3+\big(\tfrac{2}{4})^3+\big(\tfrac{3}{4})^3+\big(\tfrac{3}{4})^3=1\tag6$$
Equating terms with $(4)$ one finds $a,b,c,d,\gamma$ as,
$$a=\tfrac{7}{37},\;b=\tfrac{-4}{37},\;c=\tfrac{17}{37},\;d=\tfrac{-15}{37},\;\gamma=\tfrac{3}{4}$$
If there is a zero term,
$$0+\big(\tfrac{1}{7})^3+\big(\tfrac{1}{7})^3+\big(\tfrac{5}{7})^3+\big(\tfrac{6}{7})^3=1\tag7$$
then,
$$a=\tfrac{6}{127},\;b=\tfrac{-13}{127},\;c=\tfrac{37}{127},\;d=\tfrac{-59}{127},\;\gamma=\tfrac{6}{7}$$
and so on.
