Write $(12)(13)(14)$ in the form $\alpha(12)$ Write $(12)(13)(14)$ in the form $\alpha(12)$. Where $\alpha$ is in $A_4$.
The solution to this problem says solving $(12)(13)(14) = \alpha(12)$ for $\alpha$ we have $\alpha = (12)(13)(14)(12)$.
How come I can't solve it like this:
$(12)(13)(14)(21) = \alpha(12)(21) = \alpha$
Write $(1234)(12)(23)$, in the form $\alpha(1234)$. Where $\alpha$ is in $A_4$.
The answer is:Solving  $(1234)(12)(23) = \alpha(1234)$ for $\alpha$ we have $\alpha = (234)$.
Why can't I solve it like this $(1234)(12)(23)(4321) = \alpha(1234)(4321)$, so $(1234)(12)(23)(4321) = \alpha$

You can do that, your solution works perfectly, though you might want to evaluate them further. 
Note also that a transposition is self inverse, i.e. $(21)=(12)^{-1}=(12)$. 
Calculating $(1234)(12)(23)(4321)$, i.e. applying from right to left:
$$1\mapsto4\mapsto1 \\ 2\mapsto1\mapsto2\mapsto3\\
3\mapsto2\mapsto3\mapsto4\\
4\mapsto3\mapsto2\mapsto1\mapsto2$$
it will indeed give the cycle $(234)=(342)=(423)$ as the result.
