This is a famous Diophantine problem, to which Dickson's
*History of the Theory of Numbers, Vol. II* devotes many pages.
It is usually phrased as
w^{3}+x^{3}+y^{3}=z^{3}
or
w^{3}+x^{3}=y^{3}+z^{3},
with the implication that the variables are to be positive,
as in the integer solutions
3^{3}+4^{3}+5^{3}=6^{3}
(an amusing counterpart of the classic identity
3^{2}+4^{2}=5^{2} --
but no, it does not generalize further) and
12^{3}+1^{3}=10^{3}+9^{3}=1729
(which famously occurs in the Ramanujan taxicab story).
If we allow negative variables, the three forms of the equation
become equivalent by sign changes, and we choose the form
w^{3}+x^{3}+y^{3}+z^{3}=0
that displays the greatest symmetry. The two solutions above
are then (w,x,y,z)=(3,4,5,-6) and (x,y,z,t)=(12,1,-10,-9).
These, together with (1,6,8,-9) and variations obtained
by scaling or permuting the variables, are minimal
when we exclude trivial solutions of the form
(x,-x,y,-y) and permutations thereof.

**Theorem.**
*Suppose* (w,x,y,z)
*is a rational solution of*
w^{3}+x^{3}+y^{3}+z^{3}=0.
*Then there exist* r,s,t *such that*
(w,x,y,z) *are proportional to*

t

-t

(s-2r)t

This map (w:x:y:z) -> (r:s:t) blows down six disjoint lines, as follows:

w+az = x+ay = 0 | to | (-a:1:1); |

w+a'z = x+a'y = 0 | to | (-a':1:1); |

x+az = y+aw = 0 | to | (0:1:-a); |

x+a'z = y+a'w = 0 | to | (0:1:-a'); |

y+az = w+ax = 0 | to | (1:-a':-a); |

y+a'z = w+a'x = 0 | to | (1:-a:-a'); |

Of the fifteen lines connecting six of the blown-up points,
three are defined over **Q**. We have already seen
(s+t:s:t) mapping to (s:-t:-s:t); the other two lines are
(0:s:t), which maps to (s:-t:t:-s), and
(r:s:s), which maps to (-x:x:y:-y) with x=-(r+s), y=2r-s.

While we stated the result only for rational solutions, it holds equally over any field F not of characteristic 3, as long as F does not contain the cube roots of unity a,a'. If F does contain them, we must exclude points (w,x,y,z) on one of the six blown-down lines, and also give separate formulas for r,s,t on two further lines.

The surface w^{3}+x^{3}+y^{3}+z^{3}=0
has 4! obvious symmetries, obtained by permuting the variables.
It turns out that the even permutations preserve the set of six
blown-down lines, and thus give rise to linear transformations
of the (r:s:t) projective plane. This is not obvious from our formulas;
the relevant symmetries of **P**^{2} can be seen
more clearly after the linear change of variable

This makes our formulas longer (no coefficients vanish),
and also ill-behaved in characteristic 2; but has the advantage
that the symmetries of **P**^{2} become
monomial permutation matrices.
Thus: let F(A,B,C) be the homogeneous cubic

Then w,x,y,z are proportional to

This displays the four-group symmetry as diagonal 3*3 matrices of determinant 1. Moreover, from the pattern of coefficients of F:

The odd permutations of {w,x,y,z} send the six lines to a different set of six disjoint lines on the cubic surface, namely the set that forms a ``double six'' with the first set.