$\gcd(a^3,b^3)$ = $\gcd(a,b)^3$
Let there be integers $s,t,x,y$
$a^3s + b^3t = (ax + by)^3 $
Should I start like from the above?
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Best to consider prime factorisations instead. – πr8 Mar 01 '16 at 19:36
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If $c|d$ then $c^3|d^3$ and let c be the gcd in this case – Mar 01 '16 at 19:36
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1First prove the case where $\gcd(a,b)=1$. You can do this by taking $(ax+by)^5=a^3X+b^3Y$. Then prove the general case from there. – Thomas Andrews Mar 01 '16 at 19:43
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https://math.stackexchange.com/questions/75676/how-can-i-prove-that-gcda-b-1-implies-gcda2-b2-1-without-using-prime-d/75678#75678 – Erick Wong Mar 01 '16 at 19:43
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@ThomasAndrews how would I prove for a general case? – Ming Wu Mar 02 '16 at 01:34
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$\displaystyle ~ \gcd(a^3, b^3) = \prod_{1 \le k \le n} p_k^{\min(3r_k, 3s_k)} = \prod_{1 \le k \le n} p_i^{3\min(r_k, s_k)} = \bigg(\prod_{1 \le k \le n} p_i^{\min(r_k, s_k)} \bigg)^3 = (\gcd(a, b))^3.$
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