I want to find solutions to the equation $x^3 + 2x - 3 \equiv 0 (mod 45)$. I have already found solutions to $x^3 + 2x - 3 \equiv 0 (mod 5)$ and $x^3 + 2x - 3 \equiv 0 (mod 9)$, simply by brute force. Respectively, $x = 1, 3 (mod 5)$ and $x = 1, 2, 6 (mod 9)$ are zeros of the equations. However, the brute force method doesn't seem like a wise way to approach this problem for $mod 45$.
I'm aware of the CRT, I'm just unsure how to apply that here if applicable. This is for homework, so I'd prefer a hint or something to point me in the right direction rather than an answer.
Using CRT we can work out the general solution of $\,x\equiv a\pmod 9,\ x\equiv b\pmod 5,\,$ then specialize it for each pair $\,a,b.$
${\rm mod}\ 9\!:\ x\equiv a\,\Rightarrow\,x = a+9n.\,$ $\,{\rm mod}\ 5\!:\ b\equiv x\equiv a+9n\equiv a-n\,\Rightarrow\,n\equiv a-b\,$ therefore $\, n = a-b+5k,\,$ so $\,x = a+9n = a+9(a-b + 5n) = 10a-9b + 45k.\,$
For example if $\,x\equiv a\equiv 2\pmod 9,\,$ and $\,x\equiv b\equiv 3\pmod 5\,$ then from the above we deduce $\,x\equiv 10a-9b\equiv 10(2)-9(3)\equiv -7\equiv 38\pmod{45}.\,$
Computing $\,10a-9b\pmod{45}\,$ for all other pairs $a,b$ yields all $6$ solutions.