Question 87264
You are looking for factorised versions that have same factors in so that we can cancel down and thus simplify.


i will do the numerator first:

{{{ (5a^2-20)/(3a^2-12a)  }}}
{{{ (5(a^2-4))/(3a(a-4))  }}}
{{{ (5(a+2)(a-2))/(3a(a-4))  }}}


And now the denominator --> has multiplications and divisions so it shouldn't matter which order we do them: I shall factorise everything first to see what i have.


{{{ ( (a^3+2a^2)*(9a^3+6a^2) ) / ( (2a^2-8a)*(2a^2-4a) ) }}}
{{{ ( (a^2(a+2))*(3a^2(3a+2)) ) / ( (2a(a-4))*(2a(a-2)) ) }}}
{{{ ( (a^2(a+2))*(3a^2(3a+2)) ) / ( (2a(a-4))*(2a(a-2)) ) }}}
{{{ ( 3a^4(3a+2)(a+2) ) / ( 4a^2(a-4)(a-2) ) }}}
{{{ ( 3a^2(3a+2)(a+2) ) / ( 4(a-4)(a-2) ) }}}


what we have left now looks just as complicated but that is only because we have so many fractions in there.


Consider {{{ (a/b)/(c/d) }}}. This can be re-written as {{{ (a/b) * (d/c) }}}


so our {{{ ( (5(a+2)(a-2))/(3a(a-4)) ) / ( ( 3a^2(3a+2)(a+2) ) / ( 4(a-4)(a-2) ) ) }}} can be thought of as 


{{{ ( (5(a+2)(a-2))/(3a(a-4)) ) * ( (4(a-4)(a-2)) / (3a^2(3a+2)(a+2)) ) }}}
{{{ ( (5(a+2)(a-2)4(a-4)(a-2))/(3a(a-4)3a^2(3a+2)(a+2)) ) }}}
{{{ ( (20(a+2)(a-2)(a-4)(a-2))/(9a^3(a-4)(3a+2)(a+2)) ) }}}


and finally we can start to simplify by cancelling out terms.
{{{ ( (20(a-2)(a-4)(a-2))/(9a^3(a-4)(3a+2)) ) }}}
{{{ ( (20(a-2)(a-2))/(9a^3(3a+2)) ) }}}


which is nowhere near what you say is the answer so i am guessing that my view of what you wrote is wrong... my method is fine - i have checked it a couple of times but admittedly it is difficult on here rather than on paper.


See if you follow what i did and apply it to your version of the question and see if you get the answer you quote.


Good luck
cheers
Jon.