System of equations using the extreme principle Find the integers $a,b,c,x,y,z$ that verify the system:
$$x^3+2y^3=a^3$$
$$5y^3+3z^3=b^3$$
$$4z^3+6x^3=c^3$$
This is a logical continuation of this another problem:
Find the integers $a,b,x,y$ that verify the system:
$$x^2+6y^2=a^2$$
$$6x^3+y^2=b^2$$
You assume there is a minimal non-trivial solution $(|x|,|y|,|a|,|b|)$, Add the expressions, and because of the modulo classes of perfect squares with 7, it must be that $a,b$ are divisible by 7, so you divide by 7, and you find another even smaller solution, resulting into a contradiction.
That clearly doesn't work for the first problem, since there are some exceptions regarding the modulo classes.

I think you can also work modulo $9$ here. The cubes modulo $9$ are $-1$, $0$ and $1$. You can always swap $(a,b,c,x,y,z) \to (-a,-b,-c,-x,-y,-z)$, so you have cases $y^3 \equiv 0 \pmod 9$ or $y^3 \equiv 1 \pmod 9$.
In the first case, you get $3z^3 \pmod 9 \in \{-1,0,1\}$, so $z$ is divisible by three, and using the last equation similarly we get $x$ divisible by $3$. Thus we can reduce $(a,b,c,x,y,z) \to (a/3,b/3,c/3,x/3,y/3,z/3)$.
In the second case, $3z^3+5 \pmod 9 \in \{-1,0,1\}$, so $z^3 \equiv 1 \pmod 9$. The last equation then gives $6x^3+4 \pmod 9 \in \{-1,0,1\}$, so $x^3 \equiv 1 \pmod 9$. But then we have a contradiction in the first equation.
