A5 According to a fixed rule, we transform a triplet of numbers $(a, b, c)$ into a new triplet as follows: from $(a, b, c)$ we make $(a b, b c, c a)$. So from $(2,4,3)$ we get $(2 \times 4,4 \times 3,3 \times 2)=(8,12,6)$. We start with a triplet of different positive numbers $(a, b, c)$ and repeat this process, so $(a, b, c) \rightarrow(a b, b c, c a) \rightarrow\left(a b^{2} c, b c^{2} a, c a^{2} b\right)$ and so on...

After a number of steps, we get a triplet $(p a, p b, p c)$, where the three numbers are again in the same ratio as the initial numbers, namely $a: b: c$.

Express $p$ in terms of $a, b$, and $c$.

Note! In a triplet, the order of the numbers is important, so $(2,3,4)$ is not the same as $(2,4,3)$
A5. $(a, b, c) \rightarrow(a b, b c, c a) \rightarrow\left(a b^{2} c, b c^{2} a, c a^{2} b\right)=(b(a b c), c(a b c), a(a b c)) \rightarrow\left(b c(a b c)^{2}, c a(a b c)^{2}, a b(a b c)^{2}\right) \rightarrow$ $\left(c(a b c)^{5}, a(a b c)^{5}, b(a b c)^{5}\right) \rightarrow\left(c a(a b c)^{10}, a b(a b c)^{10}, b c(a b c)^{10}\right) \rightarrow\left(a(a b c)^{21}, b(a b c)^{21}, c(a b c)^{21}\right)$, so $p=(a b c)^{21}$.