1.Introduction

Inspired by the problem x^2+y^2=2(u^2+v^2) with the StackExcahge,

We show a parametric solution of A^3 + B^3 = n(C^3 + D^3).

This parametric solution gives infinite integral solutions for any n.

2.Theorem
     
    There is a parametric solution of A^3 + B^3 = n(C^3 + D^3).
    
    A = -p^7+3p^5c-3qp^4a-3p^3c^2+6qp^2ca+(-3q^2a^2+q^6n^2)p
    B = p^7-3p^5c+(-3nq^3+3qa)p^4+3p^3c^2+(6cnq^3-6cqa)p^2+(3q^2a^2-6q^4an+2q^6n^2)p
    C = -qp^6+3qp^2c^2+(3cnq^4-6caq^2)p+3q^3a^2-3q^5an+q^7n^2
    D = -2qp^6+6qp^4c+(3q^4n-6q^2a)p^3-3qp^2c^2+(-3cnq^4+6caq^2)p-3q^3a^2+3q^5an-q^7n^2
       
   n,a,c,p,q are arbitrary.
 
Proof.

A^3 + B^3 = n(C^3 + D^3)...................................................................(1)

Let  A=px+a, B=-px+b, C=qx+c, D=-qx+d......................................................(2)

Substitute (2) to equation (1), then we obtain

(3ap^2+3bp^2-n(3cq^2+3dq^2))x^2+(3a^2p-3b^2p-n(3c^2q-3d^2q))x+a^3+b^3-n(c^3+d^3)=0.........(3)

Equating to zero the coefficient of x^2, then we obtain

n = p^2(a+b)/(q^2(c+d))
x = -1/3(-q^2a^2+q^2ba+c^2p^2-cdp^2+d^2p^2-b^2q^2)/(qp(-qa+pc-pd+qb)).

Finally, we obtain a parametric solution. 

Q.E.D.　
 
　　                  
       
3.Example

n=2

A = -p^7+3p^5c-3qp^4a-3p^3c^2+6qp^2ca+(-3q^2a^2+4q^6)p
B = p^7-3p^5c+(-6q^3+3qa)p^4+3p^3c^2+(12cq^3-6cqa)p^2+(3q^2a^2-12q^4a+8q^6)p
C = -qp^6+3qp^2c^2+(6cq^4-6caq^2)p+3q^3a^2-6q^5a+4q^7
D = -2qp^6+6qp^4c+(6q^4-6q^2a)p^3-3qp^2c^2+(-6cq^4+6caq^2)p-3q^3a^2+6q^5a-4q^7

a,c,p,q are arbitrary.

In this way, this parametric solution gives infinite solutions.