System of equations on $\mathbb{Z}$ Knowing that $a,b,c,d,e,f \in \mathbb{Z}$ solve the system:
$$
\begin{align}
18&=cf \\
27&=ce+bf \\
28&=af+be+cd \\
22&=f+ae+bd+c \\
11&=e+b+ad \\
5&=a+d
\end{align}
$$
I want to find solution without using trail and error approach...
 A: Your system of equations basically states
$$\begin{gather}
(1+ax+bx^2+cx^3)(1+dx+ex^2+fx^3)=p(x)\\
\text{where}\quad
p(x)=1+5x+11x^2+22x^3+28x^4+27x^5+18x^6
\end{gather}$$
where $x$ is just a formal variable -- you do not solve for $x$.
So basically you have to factor the polynomial $p(x)$ into two cubic polynomials over $\mathbb{Z}$. The complete factorization into irreducibles is
$$p(x) = (1+x+2x^2+3x^3)\underbrace{(1+3x)(1+x+2x^2)}_{1+4x+5x^2+6x^3}$$
so there are only two possibilities for matching the cubic factors:
$$(a,b,c;d,e,f) \in \{(1,2,3;4,5,6),(4,5,6;1,2,3),\}$$
Normally, we would additionally have to consider the possibility of having
a common factor such as $\pm1$ distributed among the polynomial factors,
but the problem statement has constant terms fixed at $1$
which simplifies things a bit.
A: As John observed, there are $10$ different cases for $c$ and $f$. That uses up the first equation. You can then use the last and the second (now linear) equation to eliminate, say, $d$ and $e$. That leaves you with $3$ quadratic equations in $a$ and $b$. You can solve the fifth equation (now linear in $b$) for $b$ and substitute the result into the third or fourth equation to obtain a fourth or third order equation for $a$, respectively. For any integer values of $a$ that satisfy that equation (at most $4$, obtainable without trial and error), the corresponding values of $b$, $d$ and $e$ follow by substitution.
