Find triples $(a,b,c)$ of positive integers such that... Find the triples $(a,b,c)$ of positive integers that satisfy $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)=3. $$

I found this on a local question paper, and I am unable to solve it.

Any help will be appreciated.
We have $(1+\frac{1}{3})(1+\frac{1}{2})^2=3$, so that is one solution. It also shows that at least one of $a,b,c$ must be $&lt;3$. wlog we may take $a\le b\le c$. So $a=1$ or $2$.

Suppose $a=1$. Then $(1+\frac{1}{b})(1+\frac{1}{c})=\frac{3}{2}$. Since $(1+\frac{1}{5})^2&lt;\frac{3}{2}$ we must have $b&lt;5$. Obviously we need $b&gt;2$. We find $b=3$ gives the solution $(a,b,c)=(1,3,8)$ and $b=4$ gives the solution $(1,4,5)$.

Suppose $a=2$. Then we have $(1+\frac{1}{b})(1+\frac{1}{c})=2$. Since $(1+\frac{1}{3})^2&lt;2$ we must have $b&lt;3$. That gives the solution already noted of $(2,2,3)$.
Let $a=b=2$ and $c=3$.  So you have $(3/2)(3/2)(4/3)=3$.