<s>[INST] How can I find all integers $x≠3$ such that $x−3|x^3−3$ How can I find all integers $x≠3$ such that $x−3|x^3−3$?
I tried expand $x^3−3$ as a sum but I couldn't find a way after that.
 [/INST] I am sitting with this exact problem right now. I used a more brute approach to solving the problem but I think that I am missing some solutions. Could someone take a look at my solution?
Consider $$\dfrac{x^{3}-3}{x-3}=\cdots =1-\dfrac{x(x+1)(x-1)}{x-3}.$$ Therefore $$x-3\vert x^{3}-3 \Leftrightarrow x-3\vert x~~or~~x-3\vert x+1~~or~~x-3\vert x-1.$$ Now, we may consider each case by themselves and combine each solution set from these to solve the problem.
$(i)$ Consider $$\dfrac{x}{x-3}=1+\dfrac{3}{x-3}$$ and $x-3$ divides $3$ exactly when $x\in A= \{0,4,6\}$.
$(ii)$ Consider $$\dfrac{x+1}{x-3}=1+\dfrac{4}{x-3}$$ and $x-3$ divides $4$ exactly when $x\in B=\{-1,4,5,7\}$.
$(iii)$ Lastly consider $$\dfrac{x-1}{x-3}=1+\dfrac{2}{x-3}$$ and $x-3$ divides $2$ exactly when $x\in C=\{1,4,5\}$.
Answer: $x-3$ divides $x^{3}-3$ when $x\in A\cup B\cup C=\{-1,0,1,4,5,6,7\}$.
Is this correct?
</s>