1. solving cubic congruences

Find all the solutions of $x^3+3x-8\equiv 0mod(33)$

I can only see
$x^3+3x-8\equiv 0 mod(3)$ and $x^3+3x-8 \equiv 0 mod(11)$

however i cannot seem to condense these further and mod 11 by inspection does seem right

2. Originally Posted by FGT12
Find all the solutions of $x^3+3x-8\equiv 0mod(33)$

I can only see
$x^3+3x-8\equiv 0 mod(3)$ and $x^3+3x-8 \equiv 0 mod(11)$

however i cannot seem to condense these further and mod 11 by inspection does seem right
well i'll be the one who brings up thoughtless brute force, someone else could discuss more intelligently.

easiest to just show you a screenshot from pari/gp. as you can see i solved it in two ways and the answers match.

3. Originally Posted by FGT12
Find all the solutions of $x^3+3x-8\equiv 0mod(33)$

I can only see
$x^3+3x-8\equiv 0 mod(3)$ and $x^3+3x-8 \equiv 0 mod(11)$

however i cannot seem to condense these further and mod 11 by inspection does seem right
This is my attempt.
For the first congruence: $x^3\equiv x\bmod3$ by Fermat's Theorem, and replacing coefficients modulo 3 gives $x^3+3x-8\equiv x-8\equiv x+1\bmod3$ or $x\equiv-1\bmod3$.
Similarly, we replace coefficients modulo 11 and hence $x^3+3x-8\equiv 12x^3+3x+3\equiv0\bmod11$. Dividing by 3 gives $4x^3+x+1\equiv0\bmod11$. Here by simply checking integers $0\le x\le10$ gives $x\equiv5, 8, 9\bmod11$.

Now we have to consider three systems: $x\equiv-1\bmod 3$ and $x\equiv5\bmod11$; $x\equiv-1\bmod3$ and $x\equiv8\bmod11$; $x\equiv-1\bmod3$ and $x\equiv9\bmod11$. By inspection, the solutions are $x\equiv 5, 8, 20\bmod 33$ .