Question 144500
Given: {{{((x - 3)^3/2^3) * (1^2/(x-3)^2) }}}

Multiplying alegbraic fractions uses the same 'rules' as multiplication of plain old regualr fractions.

You make a simple product of the values in the numerator. Then do the same for the products of the terms in the denominator. 

Then simply (cancel out) as best you can.

Finally multiply (expand) as makes sense.

In this case:
 {{{((x - 3)^3/2^3) * (1^2/(x-3)^2) }}}

Multiply the two numerators and the two denominators, keeping their products "on their side of the division sign'.
 {{{((x - 3)^3* (1^2))/(2^3 * (x-3)^2) }}}

You can see the term {{{x-3}}} in both the numerator and the denominator. You can cancel out two of those as follows
 {{{((x - 3) * (1^2))/2^3 }}}

Now finish your simplification by finding the values for the integer powers
  {{{((x - 3) * 1)/8 }}}

Lucky you, in this case, there is nothing left to expand since the 1 can be 'canceled' 
  {{{(x - 3) /8 }}}