I have this:
$$\frac{a}{b} - \frac{a}{c} = 1$$ Solve for $c$. Then,
$$\frac{a}{b} - \frac{a}{c} = 1 \cdot bc$$
$$ = ac - ab = bc$$
$$ = a(c - b) = bc$$
$$ c = \frac{bc}{a} + b$$ This is my final result.
But the correct result is:
$$c=\frac{ab}{a-b}$$
What I development wrong in this equation ?
To isolate $c$ we can proceed as follow
$$\frac{a}{b} - \frac{a}{c} = 1\iff \frac{a}{c}=\frac{a}{b}-1=\frac{a-b}{b}\iff\frac{c}{a}=\frac{b}{a-b}\iff c=\frac{ab}{a-b}$$
You have expressed the solution to $c$ in terms of $c$, where you need to have it in terms of $a$ and $b$. This is what gives you a different(although correct) answer.
From $ac-ab=bc$, you should have taken the variables with $c$ to one side, like so:
$$ac-bc=ab \implies c(a-b)=ab$$
Can you continue from here?
Continuing from your 3rd step. $$ac-ab = bc \implies ac-bc=ab \implies c(a-b) = ab \implies c=\frac{ab}{a-b}$$
Your algebra is correct but $$c = \frac{bc}{a} + b$$ is solving $c$ in terms of $a$, $b$, and $c$ which is sort of cyclic. We want to see $c$ in terms of only $a$ and $b$.
You have to isolate $c$ by keeping all terms involving $c$ on the left side and other terms on the right side.
Note that $$ ac - ab = bc\implies c(a-b)=ab \implies c= \frac {ab}{a-b}$$
