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If $a$ divides $b$ argue that $a^3$ divides $b^5c$ for any $c$.

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If $a$ divides $b$, then $b$ can be written as: $$ b = ka $$

Therefore: $$ b^5c = (ka)^5c = a^3(k^5a^2c) $$

Which is clearly divisible by $a^3$.

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Since $a$ divides $b$, we know $a^3$ divides $b^3$ or any multiple thereof. In particular, $a^3$ divides $b^3(b^2c)$ (a multiple of $b^3$), which is $b^5c$.

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$\rm\begin{eqnarray}{\bf Hint}\quad\left(\dfrac{b}a\right)^3\! b^2 c\, &=&\rm\ \dfrac{b^5c}{a^3}\\ \rm hence\ \ \ \dfrac{b}a\,\in\, \Bbb Z\ &\Rightarrow&\rm\ \dfrac{b^5c}{a^3}\,\in\,\Bbb Z\\ \rm i.e.\qquad a\:|\:b\ &\Rightarrow&\rm\ \ a^3\:|\:b^5 c\end{eqnarray}$

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