1.Introduction

We show that ax^3+by^3=cz^3 has infinitely many integer solutions
                 
                 
2.Theorem
     
Diophantine equation ax^3+by^3=cz^3 has infinitely many integer solutions.

condition: 
a,b are arbitrary
c=a+b.

 
Proof.

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

Let x=t+1, y=-t+1, z=pt+1, c=a+b, the equation (1) becomes to below equation.

(a-b-(a+b)p^3)t^2+(3a+3b-3(a+b)p^2)t+3a-3b-3(a+b)p=0 ...................................................(2)

Since this is a quadratic equation in t, for t to be rational number, the discriminant of the equation must be square number.
Let U=p, then we obtain,
V^2 = (-3a^2-6ab-3b^2)U^4+(12a^2-12b^2)U^3+(-18a^2-36ab-18b^2)U^2+(12a^2-12b^2)U-3a^2+42ab-3b^2 ........(3)
    
Since quartic equation (3) has a rational point Q(U,V)=( (a-b)/(a+b), 12ba/((a+b) ), then this quartic equation is birationally equivalent to an elliptic curve below.

Y^2+(-8a+8b)YX = X^3+(-16a^2-40ab-16b^2)X^2+1728a^2b^2X-27648a^4b^2-69120a^3b^3-27648a^2b^4

Transformation is given, 

U = (-384a^3b-960a^2b^2+Ya-384ab^3+24baX-Yb)/(Ya+Yb)
V = (6144b^5aX+29952b^3a^3X-1440b^2a^2X^2+6144ba^5X+30720b^2a^4X+30720b^4a^2X+829440b^4a^4-576b^3aX^2-576ba^3X^2+331776b^5a^3+331776b^3a^5+12abX^3+2304a^2Yb^3-2304a^3Yb^2-1536a^4Yb+1536ab^4Y)
  /(Y^2a+Y^2b).

The point corresponding to point Q is P(X,Y)=(16a^2+40ab+16b^2, 128a^3+192a^2b-192ab^2-128b^3).

Hence we get 2Q(U,V)=((a^4-7a^3b-15a^2b^2-7ab^3+b^4)(a-b)/((a+b)(a^4+5a^3b+15a^2b^2+5ab^3+b^4)), -36(b^2+ab+a^2)(-b^3+3ab^2+6a^2b+a^3)(-b^3-6ab^2-3a^2b+a^3)ab/((a+b)(a^4+5a^3b+15a^2b^2+5ab^3+b^4)^2)  ).

This point P is of infinite order, and the multiples mP, m = 2, 3, ...give infinitely many points.

Hence we can obtain infinitely many integer solutions for equation (1).

Case : m=2

x = (2b+a)(-2b^4-4ab^3-12b^2a^2-10a^3b+a^4)
y = (b+2a)(-b^4+10ab^3+12b^2a^2+4a^3b+2a^4)
z = (a-b)(a^4+14a^3b+24b^2a^2+14ab^3+b^4)


x = -a^8-25ba^7-40b^2a^6+65b^3a^5+275b^4a^4+305b^5a^3+125b^6a^2+20b^7a+5b^8
y = 5a^8+20ba^7+125b^2a^6+305b^3a^5+275b^4a^4+65b^5a^3-40b^6a^2-25b^7a-b^8
z = -a^8+17ba^7+107b^2a^6+164b^3a^5+155b^4a^4+164b^5a^3+107b^6a^2+17b^7a-b^8

       
Q.E.D.