1.Introduction

By Tito Piezas[1],it seems that x^3+y^3+z^3 = 2w^3 has a follwing parametric solution.

A. Werebrusow : 

(p^2r^3+qs^3)^3 + (p^2r^3-qs^3)^3 + (-6rs^2)^3 = 2(p^2r^3)^3,  if pq = ±6

Generalize the above identity,I found a parametric solution of X^3 + Y^3 + Z^3 = mW^3, m>=3.

If p^3+q^3+r^3=m has one solution,then we always get a parameter solution of X^3 + Y^3 + Z^3 = mW^3.

But, the case of m =4,5 mod 9,we can't use this method.



[1].Tito Piezas:http://sites.google.com/site/tpiezas/011





2.Theorem
    
     
     
    Condition:    

    p,q,r,m:integer
    If p^3+q^3+r^3=m has a solution,then



　　There is a parameter solution of X^3 + Y^3 + Z^3 = mW^3.




     
Proof.

X^3 + Y^3 + Z^3 = mW^3....................................................(1)

X=px+a, Y=qx-a, Z=rx+b, W=x+c.............................................(2)

Substitute (2) to (1), and simplifying (1),we obtain

(3ap^2-3aq^2+3br^2-3mc)x^2+(3a^2p+3a^2q+3b^2r-3mc^2)x+b^3-mc^3=0
m=p^3+q^3+r^3

Decide c to 3ap^2-3aq^2+3br^2-3mc=0,then

c= (ap^2-aq^2+br^2)/(p^3+q^3+r^3)


x=1/3*(p^5a^3-p^5b^3+p^4qb^3-p^4qa^3-2p^3a^3q^2+3p^3a^2br^2-p^3q^2b^3-p^2b^3q^3-2p^2b^3r^3+2p^2a^3q^3-3p^2qa^2br^2-3pa^2q^2br^2+3pab^2r^4+pa^3q^4+pq^4b^3+2pqb^3r^3+3a^2br^2q^3-3qab^2r^4-a^3q^5-q^5b^3-2q^2b^3r^3)
 /((p^2a^2q+p^2b^2r+pa^2q^2-2pabr^2-pqb^2r+a^2r^3+2qabr^2+q^2b^2r)(p^3+q^3+r^3))

Substitute c and x to (2),and obtain a parameter solution.            


   
Q.E.D.　
 
　　                  
       
3.Example

I show two examples.



    
Case m=3,(p,q,r)=(1,1,1)

(18ab^2-8b^3+54a^3)^3+(-18ab^2-8b^3-54a^3)^3+(54ba^2+10b^3)^3
 = 3(18ba^2-2b^3)^3


Case m=6,(p,q,r)=(2,-1,-1)

     
(-54ba^2-108ab^2-70b^3)^3+(81ba^2+117ab^2+35b^3+27a^3)^3+(-117ab^2-81ba^2-91b^3-27a^3)^3
 = 6(-36ba^2-72ab^2-56b^3)^3