1.Introduction

It seems that {a^3+b^3+c^3 = d^3+e^3+f^3 , abc=def} has follwing parametric solutions.

Tito Piezas?[1] :
(a(ab-c^2))^3 + (b(-a^2+bc))^3 + (c(ab-c^2))^3 = (b(ab-c^2))^3 + (c(-a^2+bc))^3 + (a(-b^2+ac))^3

A. Gerardin[1] :

(p^2(p^3+2q^3))^3 + (-pq(2p^3+q^3))^3 + (q^2(p^3-q^3))^3 = (pq(p^3+2q^3))^3 + (-q^2(2p^3+q^3))^3 + (p^2(p^3-q^3))^3

Ajai Choudhry[2] :

(a(a^3-2b^3-c^3))^3 + (b(a^3+b^3+2c^3))^3 + (c(2a^3-b^3+c^3))^3 = (a(a^3+b^3+2c^3))^3 + (-b(2a^3-b^3+c^3))^3 + (-c(a^3-2b^3-c^3))^3



I found two parametric solutions of {a^3+b^3+c^3 = d^3+e^3+f^3 , abc=def} as follows.




[1].Tito Piezas:http://sites.google.com/site/tpiezas/011
[2].Ajai Choudhry,On The Solvability Of Two Simultaneous Symmetric Cubic Diophantine Equations
    With Applications To Sextic Diophantine Equations.



2.Theorem
         
     
　There are infinitely many solutions for a^3+b^3+c^3 = d^3+e^3+f^3 where abc=def.

  Case 1.

  (-(p-q^2)p)^3 + (-(pq-1)q)^3 + (p^2-q)^3 = ((-(pq-1)p)^3 + ((p^2-q)q)^3 + (-(p-q^2))^3

  Case 2.

  ((q^3+p^3-2)p)^3 + (-(-q^3+2p^3-1)q)^3 + (p^3-2q^3+1)^3 = (((p^3-2q^3+1)p)^3 + ((q^3+p^3-2)q)^3 + (-(-q^3+2p^3-1))^3

     
Proof.

a^3+b^3+c^3 -( d^3+e^3+f^3 ).....................................................(1)

Case 1.

a = px+s, b = qx+s/p, c = x+s/q   
d = px+s/q, e = qx+s, f = x+s/p..................................................(2)

Clearly, equation abc=def is satisfied.

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

(3sp^2+3sq^2/p+3s/q-3sp^2/q-3sq^2-3s/p)x^2
+(3ps^2+3s^2q/(p^2)+3s^2/(q^2)-3s^2p/(q^2)-3s^2q-3s^2/(p^2))x

x = -(pq+p+q)s/((p+q+1)pq)

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

This solution is the same as Piezas's one.


Case 2.

a = p(x+s), b = q(x+t), c = x
d = px, e = q(x+s), f = x+t......................................................(3)

Clearly, equation abc=def is satisfied.

Substitute (3) to (1), and simplifying (1),we obtain

(-3t-3q^3s+3p^3s+3q^3t)x^2
+(-3q^3s^2+3q^3t^2-3t^2+3p^3s^2)x
+p^3s^3-t^3-q^3s^3+q^3t^3

Decide t to {-3t-3q^3s+3p^3s+3q^3t=0},then

t = -s(-q^3+p^3)/(-1+q^3)

x = 1/3(p^3-2q^3+1)s/(-1+q^3)

Substitute t and x to (3),and obtain a parametric solution.  

This solution is the same as Choudhry's one.



Q.E.D.　
 
　　                  
       
3.Example



 Case 1.

 1<=p<10,q=2,s=1
 p

 1 (  3)^3 + ( -2)^3 + ( -1)^3 = ( -1)^3 +  ( -2)^3 + (  3)^3 , (  3)*( -2)*( -1) = ( -1)*( -2)*(  3)
 2 (  4)^3 + ( -6)^3 + (  2)^3 = ( -6)^3 +  (  4)^3 + (  2)^3 , (  4)*( -6)*(  2) = ( -6)*(  4)*(  2)
 3 (  3)^3 + (-10)^3 + (  7)^3 = (-15)^3 +  ( 14)^3 + (  1)^3 , (  3)*(-10)*(  7) = (-15)*( 14)*(  1)
 4 (  0)^3 + (-14)^3 + ( 14)^3 = (-28)^3 +  ( 28)^3 + (  0)^3 , (  0)*(-14)*( 14) = (-28)*( 28)*(  0)
 5 ( -5)^3 + (-18)^3 + ( 23)^3 = (-45)^3 +  ( 46)^3 + ( -1)^3 , ( -5)*(-18)*( 23) = (-45)*( 46)*( -1)
 6 (-12)^3 + (-22)^3 + ( 34)^3 = (-66)^3 +  ( 68)^3 + ( -2)^3 , (-12)*(-22)*( 34) = (-66)*( 68)*( -2)
 7 (-21)^3 + (-26)^3 + ( 47)^3 = (-91)^3 +  ( 94)^3 + ( -3)^3 , (-21)*(-26)*( 47) = (-91)*( 94)*( -3)
 8 (-32)^3 + (-30)^3 + ( 62)^3 = (-120)^3 + (124)^3 + ( -4)^3 , (-32)*(-30)*( 62) = (-120)*(124)*( -4)
 9 (-45)^3 + (-34)^3 + ( 79)^3 = (-153)^3 + (158)^3 + ( -5)^3 , (-45)*(-34)*( 79) = (-153)*(158)*( -5)



 Case 2.

 1<=p<10,q=2,s=1
 p

 1 (   7)^3 +  (  14)^3 + ( -14)^3 = ( -14)^3 + (  14)^3 +  (   7)^3 , (   7)*(  14)*( -14) = ( -14)*(  14)*(   7)
 2 (  28)^3 +  ( -14)^3 + (  -7)^3 = ( -14)^3 + (  28)^3 +  (  -7)^3 , (  28)*( -14)*(  -7) = ( -14)*(  28)*(  -7)
 3 (  99)^3 +  ( -90)^3 + (  12)^3 = (  36)^3 + (  66)^3 +  ( -45)^3 , (  99)*( -90)*(  12) = (  36)*(  66)*( -45)
 4 ( 280)^3 +  (-238)^3 + (  49)^3 = ( 196)^3 + ( 140)^3 +  (-119)^3 , ( 280)*(-238)*(  49) = ( 196)*( 140)*(-119)
 5 ( 655)^3 +  (-482)^3 + ( 110)^3 = ( 550)^3 + ( 262)^3 +  (-241)^3 , ( 655)*(-482)*( 110) = ( 550)*( 262)*(-241)
 6 (1332)^3 +  (-846)^3 + ( 201)^3 = (1206)^3 + ( 444)^3 +  (-423)^3 , (1332)*(-846)*( 201) = (1206)*( 444)*(-423)
 7 (2443)^3 + (-1354)^3 + ( 328)^3 = (2296)^3 + ( 698)^3 +  (-677)^3 , (2443)*(-1354)*( 328) = (2296)*( 698)*(-677)
 8 (4144)^3 + (-2030)^3 + ( 497)^3 = (3976)^3 + (1036)^3 + (-1015)^3 , (4144)*(-2030)*( 497) = (3976)*(1036)*(-1015)
 9 (6615)^3 + (-2898)^3 + ( 714)^3 = (6426)^3 + (1470)^3 + (-1449)^3 , (6615)*(-2898)*( 714) = (6426)*(1470)*(-1449)