Prove $\gcd(a^3,b^3)$ = $\gcd(a,b)^3$ $\gcd(a^3,b^3)$ = $\gcd(a,b)^3$
Let there be integers $s,t,x,y$
$a^3s + b^3t = (ax + by)^3 $
Should I start like from the above?

$\displaystyle ~ \gcd(a^3, b^3) = \prod_{1 \le k \le n} p_k^{\min(3r_k, 3s_k)} = \prod_{1 \le k \le n} p_i^{3\min(r_k, s_k)} = \bigg(\prod_{1 \le k \le n} p_i^{\min(r_k, s_k)} \bigg)^3 = (\gcd(a, b))^3.$
