<s>[INST] How to solve the diophantine equation:$ xa^3+yb^3=c^3$ Let $a,b,c,x,y \in \mathbb{Z}> 1$.  Any hint on how to solve of the diophantine equation $ xa^3+yb^3=c^3$? 
 [/INST] Consider following relations:
$1^3+6^3+8^3=9^3$
$3^3+4^3+5^3=6^3$
$⇒1^3+3^3+4^3+5^3+8^3=9^3$
or: $1^3 + 3^3+5^3+ 4^3(1+2^3)=9^3$
or: $(1^3 + 3^3+5^3). 1^3+ 4^3(1+2^3)=9^3$
or: $153 . 1^3 +4^3 . 9 = 9^3$
Compare this with equation:
$x a^3 + y b^3=c^3$
you find: $x=153, a=1, y=9, b=4, c=9$
It can be seen that there are infinitely many relations like $a^3+b^3+c^3=d^3$. Now if two or more terms of LHS of relation, for example $b^3$ and $c^3$ have a common factor like k we have:
$b^3+c^3= k^3[(b/k)^3+(c/k)^3]$
That is relation $a^3+b^3+c^3=d^3$ reduces to an equation like $X.A^3 +YB^3=C^3$ which indicates this equation may have many forms and solutions.
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