Prove that if $a \space | \space b$, then $a \space | \space 3b^3 - b^2 + 5b$.

Question

I'm asked to prove that if $a \space | \space b$, then $a \space | \space 3b^3 - b^2 + 5b$. That is, if $a$ divides $b$, it also divides $3b^3 - b^2 + 5b$. My text does not have a solution to this problem, as it's even-numbered, so I was looking to see if someone could please verify that my proof (direct) is correct:

1. Assume that $a \space | \space b$. Then by definition, $b = ax$ for some integer $x$.

2. $3b^3 = 3a^3x^3$

3. $-b^2 = -a^2x^2$

4. $5b = 5ax$

5. Putting it all together, $3b^3 - b^2 + 5b = 3a^3b^3-a^2x^2+5ax$

6. Factoring the right side, $3b^3 - b^2 + 5b = a(3a^2x^3-ax^2+5x)$

7. Because $a$ and $x$ are both integers, the factor by which $a$ is multiplied on the right side is also an integer.

8. By definition, then $a \space | \space 3b^3 -b^2 + 5b$.

Please let me know if this is a valid proof or if there are any flaws in my logic.

Answer 1

$$ b=ka \implies 3b^3-b^2+5b=b(3b^2-b+5)=a\cdot \left( 3kb^2-kb^2+5k\right). $$

Answer 2

Your reasoning is correct, but could be simplified a lot. You have $b = ax$; we want to show that $3b^3-b^2+5b = ay$ for some $y \in \Bbb Z$.

Pulling out a factor of $b$, we have: $$\begin{align} 3b^3-b^2+5b &= b(3b^2-b+5)\\ &= ax(3b^2-b+5)\\ &= a(3b^2x-bx+5x) \end{align}$$ so we have explicitly constructed the necessary $y$.