1.Introduction

We show that equation ax^3 + by^3 + cz^3 + dw^3 = 0 has a parametric solution.


2.Theorem
        
     
    ax^3 + by^3 + cz^3 + dw^3 = 0 has a parametric solution.

    x = (2ac^2+3ca^2+a^3)p^3+((3a^2b+6cab)q+6cad+3a^2d)p^2
      +((3bc^2+3ab^2+3b^2c)q^2+(6abd+6bdc)q+3ad^2+3dc^2+3cd^2)p
      +(-bc^2+b^3)q^3+3b^2q^2d+3bqd^2+d^3-dc^2
    
    y = (a^3-ac^2)p^3+((3ac^2+3a^2b+3ca^2)q+3a^2d)p^2+((3ab^2+6cab)q^2
      +(6cad+6abd)q+3ad^2)p+(2bc^2+b^3+3b^2c)q^3+(6bdc+3b^2d)q^2
      +(3bd^2+3cd^2+3dc^2)q+d^3-dc^2
    
    z = (-3ca^2-2a^3-ac^2)p^3+((-6a^2b-3cab)q-6a^2d-3cad)p^2
      +((-6ab^2-3cab)q^2-12abqd-3cad-6ad^2)p+(-bc^2-3b^2c-2b^3)q^3
      +(-6b^2d-3bdc)q^2+(-6bd^2-3bdc)q-2d^3-dc^2-3cd^2
    
    w = (a^3-ac^2)p^3+(3a^2d+3ca^2+3ac^2+3a^2bq)p^2
      +(3ab^2q^2+(6cab+6abd)q+6cad+3ad^2)p+(-bc^2+b^3)q^3
      +(3bc^2+3b^2c+3b^2d)q^2+(6bdc+3bd^2)q+2dc^2+3cd^2+d^3      
    
    condition: a+b+c+d=0

 
Proof.

ax^3 + by^3 + cz^3 + dw^3  = 0...........................................(1)

Put x = pt+1, y = qt+1, z = rt+1, w =t+1.................................(2)

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

(ap^3+bq^3+cr^3+d)t^3
+(3ap^2+3bq^2+3cr^2+3d)t^2
+(3ap+3bq+3cr+3d)t+a+b+c+d=0

By assumption, a+b+c+d=0.

Equating to zero the coefficient of t, then we obtain

r = -(ap+bq+d)/c
    
t = 3c(ap^2c+bq^2c+a^2p^2+2apbq+2apd+b^2q^2+2bqd+d^2+dc)
  /(-ap^3c^2-bq^3c^2+a^3p^3+3a^2p^2bq+3a^2p^2d+3apb^2q^2+6apbqd+3apd^2+b^3q^3+3b^2q^2d+3bqd^2+d^3-dc^2)
  
Substitute r and t to (2), then obtain a parametric solution.

   
Q.E.D.　
 
　　                  
       
3.Example


Case: (a,b,c,d)=(1, 1, 1, -3),  x^3 + y^3 + z^3 - 3w^3 = 0

x = 2p^3+(3q-9)p^2+(15-12q+3q^2)p-8-3q^2+9q
y = (-3+3q)p^2+(3q^2-12q+9)p-8+2q^3-9q^2+15q
z = -2p^3+(9-3q)p^2+(12q-15-3q^2)p-15q+10-2q^3+9q^2
w = (-1+q)p^2+(3+q^2-4q)p-q^2+3q-2