Let $a$ and $b$ be two integers. Is it true that $\operatorname{lcm}(a^2, ab, b^2) = \operatorname{lcm}(a^2, b^2)$ ?
$a, b$ are integers and the result that i tried to use, in this case,is $ \operatorname{lcm}(a^2,b^2)\gcd(a^2,b^2)=a^2 b^2$ but i don't know if it's true $\operatorname{lcm}(a^2,ab,b^2)\gcd(a^2,ab,b^2)=a^3 b^3$
I did some examples apparently it's true but I can't prove it. So could give me a hint?
Best Answer
Note $\,a^2,b^2\mid m\Rightarrow(ab)^2\!\mid m^2\Rightarrow \color{#c00}{ab}\mid m,\,$ by $\,r\in\Bbb Q,\,r^2\in\Bbb Z\Rightarrow r\in \Bbb Z,\,$ by Rational Root Test
Thus $\,\color{#0a0}{a^2,b^2}\mid m\!\iff a^2,\color{#c00}{ab},b^2\mid m^{\phantom{|^|}}\!$
so $\,\ {\rm lcm}(\color{#0a0}{a^2,b^2}) = {\rm lcm}(a^2,ab,b^2)^{\phantom{|^|}}\! $ via LCM Universal Property (or by: above $(\!\!\iff\!\!)$ shows both sides have same set $S$ of common multiples $m,\,$ so the same least common multiple $= \min S$).
Alternatively with $[x,y]:={\rm lcm}(x,y)\,$ & applying LCM distributive law to expand products
$ [a,b][aa,bb] = [aaa,aab,abb,bbb] = [a,b]^3\,$ so $\,[aa,bb] = [a,b]^2\,$ by cancelling $[a,b]\neq 0,\,$
same as in the GCD Freshman's Dream $\,(a^n,b^n) = (a,b)^n = (a^n, a^{n-1}b,\ldots, a b^{n-1}a, b^n)$