Let $a$, $b$, and $c$ be positive integers. Prove that
$$ \gcd (a, b, c) × [ab, ac, bc] = a × b × c $$
Hint: You may find the result of Exercises 1.3, question 19 useful: $\gcd (a, b, c) = \gcd (a, \gcd (b, c))$.
We start with $ \gcd (a, b, c)$
$$ \gcd (a, b, c) = \gcd (\gcd (a, b), c) \quad \text{using hint}$$
Next we consider $[ab, ac, bc]$
$$ \begin{align} [ab, ac, bc] & = [ab, [ac, bc]] \quad \text{Proposition 2.23} \\ \\ &= [ab, [a,b] \times c] \\ \\ & = [\gcd(a,b) \times [a,b], [a,b] \times c] \quad \text{Proposition 2.22} \\ \\ & = [a,b] \times [\gcd(a,b) , c] \\ \\ & = \frac{a \times b}{\gcd(a,b)} \times [\gcd(a,b) , c] \\ \\ & = \frac{a \times b}{\cancel{\gcd(a,b)}} \times \frac{\cancel{\gcd(a,b)} \times c}{\gcd( \gcd(a,b) ,c)} \\ \\ &= \frac{a \times b \times c}{\gcd( \gcd(a,b) ,c)} \end{align}$$
The product $ \gcd (a, b, c) × [ab, ac, bc] $ is
$$ \begin{align} \gcd (a, b, c) × [ab, ac, bc] & = \cancel{\gcd (\gcd (a, b), c)} \times \frac{a \times b \times c}{\cancel{\gcd (\gcd (a, b), c)}} \\ \\ & = a \times b \times c \end{align} $$