<s>[INST] $a^2 = 2b^3 = 3c^5$ Find the smallest value of $abc$. We have  following equation:
$a^2 = 2b^3 = 3c^5$
Where $a, b, c$ are natural numbers.
Find the smallest possible value  of product $abc$.
 [/INST] Introducing any prime factors apart from $2$ and $3$ makes things worse. Therefore we may assume $a=2^\alpha3^\beta$, which leads to
$$b^3=2^{2\alpha-1}3^{2\beta},\qquad c^5=2^{2\alpha}3^{2\beta-1}\ .$$
therefore we have to find the smallest $\alpha\geq0$, $\beta\geq0$ such that
$$2\alpha-1=0\quad(3),\qquad 2\alpha=0\quad(5)\ ,$$
and
$$2\beta=0\quad(3),\qquad2\beta-1=0\quad(5)\ .$$
The obvious solutions are $\alpha=5$, $\>\beta=3$, and lead to the triple $(a,b,c)=(864,72,12)$, so that the minimal product is $abc=746\,496$. 
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