<s>[INST] System of 3 linear congruences 
Find all solutions: $$\begin{cases} x\equiv 39 \mod(189) \\ x\equiv 25 \mod(539) \\ x\equiv 399 \mod(1089) \end{cases}$$

But $189=3^3\cdot7$, $539=11\cdot7^2$ and $1089=3^2\cdot 11^2$, so I can't use here Chinese remainder theorem.
 [/INST] You can find a system of congruences equivalent to the original system, but with pairwise relatively prime moduli.
For example, the first congruence is equivalent to the system $x\equiv 39\pmod{3^3}$, $x\equiv 39\pmod{7}$. 
The second original congruence is equivalent to the system $x\equiv 25\pmod{11}$ $x\equiv 25\pmod{7^2}$. 
Note that from the first congruence we had $x\equiv 4\pmod{7}$. If $ \equiv 25\pmod{7^2}$ then $x\equiv 4\pmod{7}$, our first two original congruences are equivalent to the system $x\equiv 39\pmod{3^3}$, $x\equiv 25\pmod{7^2}$,  $x\equiv 25\pmod{11}$.
The third original congruence can be similarly dealt with. When we are through we end up with a system with moduli $3^3$, $7^2$, and $11$. 
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