1.Introduction


We show a parametric solution for ax1^3 + bx2^3 + cx3^3 = ay1^n + by2^n + cy3^n.


2.Theorem
      
 
There is a parametric solution of ax1^3 + bx2^3 + cx3^3 = ay1^n + by2^n + cy3^n,
    x1 = -(2b^2(q^(1/3M))^6ap^(1/3M)+3b(q^(1/3M))^3a^2(p^(1/3M))^4+6b(q^(1/3M))^3a(p^(1/3M))^2c+3b(q^(1/3M))^3c^2+3cb^2(q^(1/3M))^6+a^3(p^(1/3M))^7+3a^2(p^(1/3M))^5c+3a(p^(1/3M))^3c^2+p^(1/3M)c^3-p^(1/3M)cb^2(q^(1/3M))^6)(ab^2(q^(1/3M))^6-a^3(p^(1/3M))^6-3a^2(p^(1/3M))^4c-3a(p^(1/3M))^2c^2-c^3+cb^2(q^(1/3M))^6)^(1/3M-1)    
    x2 = q^(1/3M)(3a^2(p^(1/3M))^3b(q^(1/3M))^3+2a^3(p^(1/3M))^6+6a^2(p^(1/3M))^4c+6a(p^(1/3M))^2c^2+3b(q^(1/3M))^3a(p^(1/3M))^2c+3cap^(1/3M)b(q^(1/3M))^3+2c^3+3b(q^(1/3M))^3c^2+ab^2(q^(1/3M))^6+cb^2(q^(1/3M))^6)(ab^2(q^(1/3M))^6-a^3(p^(1/3M))^6-3a^2(p^(1/3M))^4c-3a(p^(1/3M))^2c^2-c^3+cb^2(q^(1/3M))^6)^(1/3M-1)
    x3 = -(3b^2(q^(1/3M))^6ap^(1/3M)+3b(q^(1/3M))^3a^2(p^(1/3M))^4+6b(q^(1/3M))^3a(p^(1/3M))^2c+3b(q^(1/3M))^3c^2+2cb^2(q^(1/3M))^6-ab^2(q^(1/3M))^6+a^3(p^(1/3M))^6+3a^2(p^(1/3M))^4c+3a(p^(1/3M))^2c^2+c^3)(ab^2(q^(1/3M))^6-a^3(p^(1/3M))^6-3a^2(p^(1/3M))^4c-3a(p^(1/3M))^2c^2-c^3+cb^2(q^(1/3M))^6)^(1/3M-1)
    y1 = p^(M/n)(ab^2(q^(1/3M))^6-a^3(p^(1/3M))^6-3a^2(p^(1/3M))^4c-3a(p^(1/3M))^2c^2-c^3+cb^2(q^(1/3M))^6)^(M/n)
    y2 = q^(M/n)(ab^2(q^(1/3M))^6-a^3(p^(1/3M))^6-3a^2(p^(1/3M))^4c-3a(p^(1/3M))^2c^2-c^3+cb^2(q^(1/3M))^6)^(M/n)
    y3 = (ab^2(q^(1/3M))^6-a^3(p^(1/3M))^6-3a^2(p^(1/3M))^4c-3a(p^(1/3M))^2c^2-c^3+cb^2(q^(1/3M))^6)^(M/n)
     
    a, b, c, p, q, n: arbitrary
    M=lcm(3,n)


     
Proof.

ax1^3 + bx2^3 + cx3^3 = ay1^n + by2^n + cy3^n.........................................(1)
Let M=lcm(3,n).
Let x1=t+p^(M/3), x2=mt+q^(M/3), x3=t+1,y1=p^(M/n), y2=q^(M/n), y3=1..................(2)

(-a-bm^3-c)t^3+(-3ap^(1/3M)-3bq^(1/3M)m^2-3c)t^2+(-3a(p^(1/3M))^2-3b(q^(1/3M))^2m-3c)t=0

Then we obtain m = -(a(p^(1/3M))^2+c)/(b(q^(1/3M))^2) and,
t = -3(ap^(1/3M)b(q^(1/3M))^3+a^2(p^(1/3M))^4+2a(p^(1/3M))^2c+c^2+cb(q^(1/3M))^3)b(q^(1/3M))^3/(ab^2(q^(1/3M))^6-a^3(p^(1/3M))^6-3a^2(p^(1/3M))^4c-3a(p^(1/3M))^2c^2-c^3+cb^2(q^(1/3M))^6).

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

 
Q.E.D.
　
　
　
3.Examples

Case. ax1^3 + bx2^3 + cx3^3 = ay1^5 + by2^5 + cy3^5

x1 = (-ab^2q^30+a^3p^30+3a^2p^20c+3ap^10c^2+c^3-cb^2q^30)^4(2b^2q^30ap^5+3bq^15a^2p^20+6bq^15ap^10c+3bq^15c^2+3cb^2q^30+p^35a^3+3p^25a^2c+3p^15ac^2+p^5c^3-p^5cb^2q^30)
x2 = -(-ab^2q^30+a^3p^30+3a^2p^20c+3ap^10c^2+c^3-cb^2q^30)^4q^5(3a^2p^15bq^15+2a^3p^30+6a^2p^20c+6ap^10c^2+3bq^15ap^10c+3cap^5bq^15+2c^3+3bq^15c^2+ab^2q^30+cb^2q^30)
x3 = (-ab^2q^30+a^3p^30+3a^2p^20c+3ap^10c^2+c^3-cb^2q^30)^4(3b^2q^30ap^5+3bq^15a^2p^20+6bq^15ap^10c+3bq^15c^2+2cb^2q^30-ab^2q^30+a^3p^30+3a^2p^20c+3ap^10c^2+c^3)
y1 = p^3(-ab^2q^30+a^3p^30+3a^2p^20c+3ap^10c^2+c^3-cb^2q^30)^3
y2 = q^3(-ab^2q^30+a^3p^30+3a^2p^20c+3ap^10c^2+c^3-cb^2q^30)^3
y3 = (-ab^2q^30+a^3p^30+3a^2p^20c+3ap^10c^2+c^3-cb^2q^30)^3

Case. ax1^3 + bx2^3 + cx3^3 = ay1^6 + by2^6 + cy3^6

x1 = (ab^2q^12-a^3p^12-3a^2p^8c-3ap^4c^2-c^3+cb^2q^12)(-2b^2q^12ap^2-3bq^6a^2p^8-6bq^6ap^4c-3bq^6c^2-3cb^2q^12-p^14a^3-3p^10a^2c-3p^6ac^2-p^2c^3+p^2cb^2q^12)
x2 = (ab^2q^12-a^3p^12-3a^2p^8c-3ap^4c^2-c^3+cb^2q^12)q^2(3a^2p^6bq^6+2a^3p^12+6a^2p^8c+6ap^4c^2+3bq^6ap^4c+3cap^2bq^6+2c^3+3bq^6c^2+ab^2q^12+cb^2q^12)
x3 = -(ab^2q^12-a^3p^12-3a^2p^8c-3ap^4c^2-c^3+cb^2q^12)(3b^2q^12ap^2+3bq^6a^2p^8+6bq^6ap^4c+3bq^6c^2+2cb^2q^12-ab^2q^12+a^3p^12+3a^2p^8c+3ap^4c^2+c^3)
y1 = p(ab^2q^12-a^3p^12-3a^2p^8c-3ap^4c^2-c^3+cb^2q^12)
y2 = q(ab^2q^12-a^3p^12-3a^2p^8c-3ap^4c^2-c^3+cb^2q^12)
y3 = ab^2q^12-a^3p^12-3a^2p^8c-3ap^4c^2-c^3+cb^2q^12