$(a,b)=d \overset{?}{\implies} (a^3,b^3)=d^3$ Why is this true? I suspect that its because $\frac{LCM(a,b)^3GCD(a,b)^3}{b^3}=a^3$ and $\frac{LCM(a,b)^3GCD(a,b)^3}{a^3}=b^3$, so it must be the case for $LCM(a,b) \notin R(a,b)$, right?
 A: One way to see this is the following characterization of $(a,b)$.
Let $a=p_1^{e_1}\ldots p_n^{e_n}$ and $b=p_1^{f_1}\ldots p_n^{f_n}$ where we allow some $e_i$, $f_i$ to be zero so that we can consider the same set of primes $p_1,\ldots,p_n$.  Then
$$(a,b)=p_1^{\min\{e_1,f_1\}}\ldots p_n^{\min\{e_n,f_n\}}$$
The result follows from the observation that $3\min\{e_i,f_i\}=\min\{3e_i,3f_i\}$.
A: Using basic gcd laws (commutative, associative, distributive, etc) we have
$$\begin{eqnarray}\rm (a,b)^6\! &=\,&\rm (a^6, a^5b,a^4b^2,a^3b^3,a^2b^4,ab^5,b^6) \\
          &=\,&\rm (a^3\color{#c00}{(a^3,a^2b,ab^2,b^3), (a^3,a^2b,ab^2,b^3)}b^3) \\
          &=\,&\rm (a^3\color{#c00}{(a,b)^3,(a,b)^3}b^3) \\
          &=\,&\rm (a^3,b^3)\, \color{#c00}{(a,b)^3}  \\
\rm\Rightarrow\ \ (a,b)^3\!  &=&\rm (a^3,b^3)\ \  via\ \ cancel\ \ \color{#c00}{(a,b)^3}\end{eqnarray}$$
Remark $\ $ Similarly  $\rm\ (a,b)^n = (a^n,b^n),\ $ the Freshman's Dream for gcds. For further discussion see this post, and this post.
There are a few other ways to prove it, e.g. using prime factorizations, or using Bezout's identity, and/or factoring out the gcd to reduce to case $\rm\:(a,b) = 1,\:$ then applying Euclid's Lemma. The advantage of the above proof is that it works much more generally, e.g. as the first link shows, it generalizes to invertible ideals, and it also works in any gcd domain (where proofs employing primes need not work, since there are gcd domains with no irreducibles, so no primes, e.g. the ring of all algebraic integers).
A: Let $d=(a,b)$ then $a=da', b=db'$ and $(a',b')=1$.
Then
$$(a^n,b^n)=(d^na'^n, d^nb'^n)=d^n (a'^n,b'^n)$$
It is very easy to prove that $(a'^n,b'^n)$ cannot be divisible by any prime, thus it must be 1. 
A: Hint: Consider the prime factorizations of $a$ and $b$. The primes in common yield a product $d$. Now raise all primes to the third power.
