1.Introduction

We show a parametric solution of x1^6 + x2^6 + x3^6 = y1^3 + y2^3 + y3^3.


2.Theorem
      
 
There is a parametric solution of x1^6 + x2^6 + x3^6 = y1^3 + y2^3 + y3^3,

      x1 = a(-2c^12+a^12+3a^8b^4+3a^4b^8+b^12)
      x2 = b(-2c^12+a^12+3a^8b^4+3a^4b^8+b^12)
      x3 = c(-2c^12+a^12+3a^8b^4+3a^4b^8+b^12)
      y1 = (-2c^12+a^12+3a^8b^4+3a^4b^8+b^12)(c^12a^2+3c^12b^2+3c^6a^8+6c^6a^4b^4+3c^6b^8+a^14+3a^10b^4+3a^6b^8+a^2b^12)
      y2 = (-2c^12+a^12+3a^8b^4+3a^4b^8+b^12)(3c^12a^2+c^12b^2+3c^6a^8+6c^6a^4b^4+3c^6b^8+b^2a^12+3b^6a^8+3b^10a^4+b^14)
      y3 = -(-2c^12+a^12+3a^8b^4+3a^4b^8+b^12)c^2(3a^6c^6+3a^4b^2c^6+2a^12+6a^8b^4+6a^4b^8+3b^4a^2c^6+3b^6c^6+2b^12+2c^12)
    
     a, b, c: arbitrary.

     
Proof.

x1^6 + x2^6 + x3^6 = y1^3 + y2^3 + y3^3.............................(1)

Set x1=a, x2=b, x3=c, y1=t+a^2, y2=t+b^2, y3=mt+c^2.................(2)

(2+m^3)t^3+(3a^2+3b^2+3c^2m^2)t^2+(3a^4+3b^4+3c^4m)t=0

Then we obtain m = -(a^4+b^4)/(c^4) and t = 3(a^2c^6+b^2c^6+a^8+2a^4b^4+b^8)c^6/(-2c^12+a^12+3a^8b^4+3a^4b^8+b^12).

Substitute m and t to (2), and we obtain a parametric solution.

 
Q.E.D.