Think about the quadratic formula!

We recognise $b^2-4ac$ as the discriminant of the quadratic $ax^2+bx+c=0$ (and permutations of these). For a quadratic with integer coefficients, if the discriminant is the square of an integer, then the quadratic has rational roots, and vice-versa.

So we just need to make sure each of the quadratics $ax^2+bx+c=0$ , $bx^2+cx+a=0$ and $cx^2+ax+b=0$ has rational roots.

The sum of the roots of a quadratic with integer coefficients will always be rational. So if one of the roots is rational, the other one must be too.

All we need to do is ensure one of the roots of each of the quadratics is rational. But this is easy - just take $a+b+c=0$ , and then $x=1$ will be a root of all three of the equations. So any triple of non-zero integers satisfying $a+b+c=0$ will work, and there are clearly infinitely many of these.

Incidentally, these are not the only triples that work - an example whose sum is not zero is $(a,b,c)=(-85,9,36)$ . I'd be interested if anyone could classify ALL valid triples!

one can explore and see that for $a=-4,b=c=2$ we get an integer for the large expression. Then one needs to notice that if $(a,b,c)$ gives an integer for the expression, then so does $(ta,tb,tc)$ , where $t$ is a non-zero integer.