When can I reduce a fraction? i am currently working on a mathematics exercise out of my book.
What I basically can not get into my head yet, is when I am allowed to reduce a fraction and when not. In german there is a saying going like " out of sums/differences only the dumb reduce" (it makes sense in german as it would then rhyme). 
For example
$$\frac{ 6a^{ 4 } b^{ 2 }c^{ 2 }}{6a^{ 4 }bc-9a^{ 4 }b  }$$
results into 
$$\frac{ 6bc^{ 2 } }{6c-9  }$$
As you can see $$a^{ 4 }$$ has been reduced.
Would it now be possible to reduce the fraction further by $c$?
My book and an online algebra calculator gives me the following end result:
$$\frac{2bc^2  }{2c-3  }$$
Why can't I now reduce the fraction by $c$? I have looked up some german math pages and video on this topic already, they would say it is because $$2c-3$$ is a difference. But isn't $$6a^{ 4 }bc-9a^{ 4 }b$$ a difference just as well? In general: when can I reduce a fraction and when not? Would be nice if you could include pages that explain this topic in general, I did not find anything covering this topic specifically. 
Thanks.
 A: It helps to understand exactly what "reducing" means and why it works. Note that multiplying any number by $1$ does not change its value. When you reduce all you're doing is multiplying the fraction by another fraction equivalent to $1$. Observe,
$$\frac{ 6a^{ 4 } b^{ 2 }c^{ 2 }}{6a^{ 4 }bc-9a^{ 4 }b  }\cdot\frac{1/(3a^4b)}{1/(3a^4b)}=\frac{2bc^2}{2c-3}$$
since $\frac{1/(3a^4b)}{1/(3a^4b)}=1$ (assuming $3a^4b\neq0$). Do you see why this works?
However, if you try to multiply by $\frac{1/c}{1/c}$ it doesn't work nicely:
$$\frac{2bc^2}{2c-3}\cdot\frac{1/c}{1/c}=\frac{2bc}{2-3/c}$$
Of course this is a valid equation (assuming $c\neq0$), but fractions within fractions are kind of ugly and usually not what you want.
You can cancel nicely if the thing you're cancelling by is a factor of all terms in the numerator and denominator, if they are a sum or difference.
A: HINT:
You can reduce factors of a product. So
$$
\frac{2a^3}{6a^2}=\frac{2\cdot a\cdot a\cdot a}{2\cdot 3\cdot a\cdot a}=
\frac{\color{red}{2}\cdot a\cdot \color{red}a\cdot \color{red}a}{\color{red}2\cdot 3\cdot \color{red}a\cdot \color{red}a}=\frac{a}{3}
$$
This was easy since it was easy to recognize the factors.
A more complicated example would be
$$
\frac{8a^3}{6a^2+3a^3}
$$
since you do not have a product in the denominator rather a sum of two products BUT we can factorize the denominator in the following way, so we will only concentrate on the denominator now
$$
6a^2+3a^3=2\cdot \color{red}3\cdot \color{red}a\cdot \color{red}a+\color{red}3\cdot \color{red}a\cdot \color{red}a\cdot a=
$$
the red coloured parts are common so we can write our expression as
$$
=3\cdot a\cdot a\cdot (2+a)=3a^2(2+a)
$$
and looking at the whole fraction now
$$
\frac{8a^3}{3a^2(2+a)}=\frac{2\cdot2\cdot 2\cdot \color{red}a\cdot \color{red}a\cdot a}{3\cdot \color{red}a\cdot \color{red}a\cdot(2+a)}=
$$
where once again the red parts are factors of the nominator and denominator so they may be cancelled out resulting in
$$
\frac{8a}{3(2+a)}.
$$
Anouther example may be
$$
\frac{a^2-b^2}{a+b}=\frac{(a+b)\cdot(a-b)}{a+b}=\frac{\color{red}{(a+b)}\cdot(a-b)}{\color{red}{(a+b)}}=
$$
where again the red parts are factors of the nominator and denominator so they may be cancelled resulting in
$$
\frac{a-b}{1}=a-b
$$
where that $1$ comes from? Well it was actually there all the time since
$$
x=x\cdot 1=x\cdot 1\cdot1...
$$
so we actually had
$$
\frac{(a+b)\cdot(a-b)}{(a+b)\cdot 1}
$$
all along.
My answer is not that theoretical rather of the practical sort since I assumed it will help you the most
Hope this helped clear up some things
EDIT
Looking at your specific fraction
$$
\frac{6bc^2}{6c-9}
$$
we can see that, once again looking at only the denominator 
$$
6c-9=2\cdot \color{red}3\cdot c-\color{red}3\cdot 3=
$$
we see that a $3$ is common so we can factorize the difference as
$$
=3\cdot (2c-3)
$$
and looking at the whole fraction again
$$
\frac{6bc^2}{6c-9}=\frac{2\cdot \color{red}3\cdot b\cdot c\cdot c}{\color{red}3\cdot(2c-3)}=
$$
we see that the $3$s cancel out so you have
$$
=\frac{2bc^2}{2c-3}
$$
and now we cannot factorize the denominator more and since we can cancel only factors of a product we won't be able to cancel more.
A: Recall that when multiplying two fractions, we multiply the numerators and multiply the denominators as follows:
$$\frac{a}{c}\cdot\frac{b}{d} = \frac{ab}{cd}$$
But because equality is transitive, it is also true that we can take a fraction and split it into two or more fractions, essentially unmultiplying it.
$$\frac{ab}{cd} = \frac{a}{c}\cdot\frac{b}{d}$$
Using this idea on the example you gave we could proceed as
$$\begin{align}
\frac{6a^4b^2c^2}{6a^4bc - 9a^4b} &= \frac{(3a^4b)(2bc^2)}{(3a^4b)(2c-3)}\\
&= \frac{3a^4b}{3a^4b}\cdot\frac{2bc^2}{2c-3}\\
&= 1\cdot\frac{2bc^2}{2c-3}\\
&= \frac{2bc^2}{2c-3}\\
\end{align}$$
The reason why this can't be reduced any further is because the numerator and denominator need to have a common factor, and furthermore, that common factor also needs to be a common factor of each of the terms of the difference in the denominator [otherwise we wouldn't be able to unmultiply the expression]. Initially we saw that $6a^4b^2c^2$, $6a^4bc$, and $9a^4b$ were all divisible by $3a^4b$, but that only the first two terms were divisible by $c$. Because of this, $c$ wasn't a factor of the entire denominator.
