TITLE: Find all integer solutions to Diophantine equation $x^3+y^3+z^3=w^3$
QUESTION [7 upvotes]: Compute all integer solutions to the equation
$$x^3+y^3+z^3=w^3$$

REPLY [3 votes]: For Diophantine equations:  $X^3+Y^3+Z^3=R^3$
Symmetric solutions can be written as: $\left|\frac{X+Y}{R-Z}\right|=\frac{b^2}{a^2}$
Solutions have the form:
$X=b(a^6-b^6)p^3+3ba^6p^2s+3ba^6ps^2+b(a^6-b^6)s^3$
$Y=(a^7-ab^6)p^3+3(a^7-a^4b^3-ab^6)p^2s+3(a^7-2a^4b^3)ps^2+(a^7-3a^4b^3+2ab^6)s^3$
$Z=(2ba^6+3a^3b^4+b^7)p^3+3(2ba^6+a^3b^4)p^2s+3(2ba^6-a^3b^4)ps^2+(2ba^6-3a^3b^4+b^7)s^3$
$R=(a^7+3a^4b^3+2ab^6)p^3+3(a^7+2a^4b^3)p^2s+3(a^7+a^4b^3-ab^6)ps^2+(a^7-ab^6)s^3$
Solutions have the form:
$X=3(b^6-7a^6)as^2-9a^4ps-ap^2$
$Y=(6ab^6+21a^7-27b^3a^4)s^2-3(2ab^3-3a^4)ps+ap^2$
$Z=(3b^7+33ba^6-18a^3b^4)s^2+3(4ba^3-b^4)ps+bp^2$
$R=(3b^7+6ba^6-9a^3b^4)s^2+3(2ba^3-b^4)ps+bp^2$
Solutions have the form:
$X=(b^6-7a^6)ap^2+9a^4ps-3as^2$
$Y=(2ab^6+9b^3a^4+7a^7)p^2-3(2ab^3+3a^4)ps+3as^2$
$Z=(3b^4a^3+2ba^6+b^7)p^2-3(2ba^3+b^4)ps+3bs^2$
$R=(6b^4a^3+11ba^6+b^7)p^2-3(4ba^3+b^4)ps+3bs^2$
Solutions have the form:
$X=(aj-bt)^2sp^6-3t^2j^2p^2s^5+3(aj+bt)jtps^6-(a^2j^2+abtj+b^2t^2)s^7$
$Y=2(aj-bt)^2sp^6-6(aj-bt)jtp^4s^3+3(a^2j^2-b^2t^2)p^3s^4+3t^2j^2p^2s^5-3(aj+bt)tjps^6+$ 
$+(a^2j^2+abtj+b^2t^2)s^7$
$Z=(aj-bt)^2p^7-3(aj-bt)tjp^5s^2+3(aj-bt)btp^4s^3+3t^2j^2p^3s^4-6bjt^2p^2s^5-$
$-(a^2j^2-2abtj-2b^2t^2)ps^6$
$R=(aj-bt)^2p^7-3(aj-bt)tjp^5s^2+3(aj-bt)ajp^4s^3+3t^2j^2p^3s^4-6atj^2p^2s^5-$
$-(b^2t^2-2abtj-2a^2j^2)ps^6$
$a,b,j,t,p,s$ - what some integers.