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.
 A: Hint $\ {\rm mod}\ x\!-\!3\!:\ x\equiv 3\,\Rightarrow\, x^3\!-3\equiv 24,\ $ so $\,x\!-\!3\mid x^3\!-3\iff x\!-\!3\mid 24$
If modular arithmetic is unfamilar, by the Factor Theorem, $\, x\!-\!3\mid f(x)\!-\!f(3)\,$ so for $\,f(x) = x^3\,$ we infer $\,x\!-\!3\mid x^3\!-3^3 = (x^3\!-3)-24,\,$ thus $\,x\!-\!3\mid x^3\!-3\iff x\!-\!3\mid 24.$
Generally $\ a\mid b \iff a\mid (b\ {\rm mod}\ a),\ $ so we can often simplify divisibility statements by reducing the dividend modulo the divisor. Above $\ x^3\!-3\ {\rm mod}\ x\!-\!3\,=\, 24,\,$ which is a special case of the Polynomial Remainder Theorem $\,f(x)\ {\rm mod}\ x\!-\!a\, =\, f(a)$
Or, equivalently, we can employ the $\rm\color{#c00}{EA} = $ Euclidean Algorithm for the gcd as follows
$$ a\mid b\iff a = (a,b)\overset{\rm\color{#c00}{EA}} = (a,\, b\bmod a)\iff a\mid (b\bmod a)$$
A: $$x - 3 | x^3 - 3$$
$$\frac{x^3 - 3}{x - 3} = \frac{x^3 - 3 + 27 - 27}{x - 3}$$
$$ = \frac{x^3 - 27 + 24}{x - 3}$$
$$ = \frac{x^3 - 3^3 + 24}{x - 3}$$
$$ = \frac{(x - 3) \times (x^2 + 3x + 9) + 24}{x - 3}$$
$$ = x^2 + 3x + 9 + \frac{24}{x - 3}$$
The result is integer when $x - 3$ is a factor of $24$ and $x \ne 3$
A: 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?
