Integer solutions of $ x^3+y^3+z^3=(x+y+z)^3 $ Consider the equation
$$ x^3+y^3+z^3=(x+y+z)^3 $$
for triples of integers $(x, y, z) $.
I noticed that this has infinitely many solutions: $ x, y $ arbitrary and $ z=-y $. 
Are there more solutions?
 A: Too long for a comment. Interestingly, for $k>3$, there are non-trivial solutions to,
$k=4;\quad x_i =  -2634, 955, 1770, 5400$: (Jacobi-Madden equation)
$$x_1^4+x_2^4+x_3^4+x_4^4 = (x_1+x_2+x_3+x_4)^4$$
$k=5;\quad x_i =  - 3, - 54,24,28,67 $:
$$x_1^5+x_2^5+\dots+x_5^5 = (x_1+x_2+\dots+x_5)^5$$
$k=6;\quad x_i =  -4170, -3187, -888, 1854, 3300, 3936, 4230$:
$$x_1^6+x_2^6+\dots+x_7^6 = (x_1+x_2+\dots+x_7)^6$$
$k=7;\quad x_i =  -230, -353, -625, -673, 184, 443, 556$:
$$x_1^7+x_2^7+\dots+x_7^7 = (x_1+x_2+\dots+x_7)^7$$
and longer ones for $k = 8,9,10$ (See April 6 update).
P.S. The cases $k=4,5$ involve elliptic curves, hence there are an infinite number of co-prime solutions.
A: $$(x+y+z)^3-(x^3+y^3+z^3)=3(x+y)(y+z)(z+x)$$
so the only solutions are the ones the OP observed and their cyclically symmetric counterparts.  There's no essential number theory going on here, just an algebraic identity.
A: For such equations :
$$(x+y)^3+x^3+y^3+z^3=(x+y+z)^3$$
You can write the formula.
$$x=3p^2+18ps-s^2$$
$$y=15p^2-6ps-5s^2$$
$$z=3p^2-6ps+7s^2$$
