Question 975631
<pre>

{{{(3^(q+3) - 3^2(3^q))/(3(3^(q+4)))}}}

We add the exponents of 3 in the SECOND terms on top, which are 2 and q,
we get:

{{{(3^(q+3) - 3^(2+q))/(3(3^(q+4)))}}}

Change the 3 in the bottom to 3<sup>1</sup>

{{{(3^(q+3) - 3^(2+q))/(3^1(3^(q+4)))}}}

Then add the exponents of 3 in the bottom:

{{{(3^(q+3) - 3^(2+q))/(3^(q+5))}}}

Then make this into two fractions:

{{{(3^(q+3))/(3^(q+5))}}}{{{""-""}}}{{{(3^(2+q))/(3^(q+5))}}}

Subtract exponents of 3 in the fractions:

{{{3^((q+3)-(q+5))}}}{{{""-""}}}{{{3^((2+q)-(q+5))}}}

Simplify the exponents:

{{{3^(q+3-q-5)}}}{{{""-""}}}{{{3^(2+q-q-5)}}}

{{{3^(-2)}}}{{{""-""}}}{{{3^(-3)}}}

Get rid of the negative exponents by putting them in the
denominators with positive exponents:

{{{1/3^2}}}{{{""-""}}}{{{1/3^3}}}

{{{1/9}}}{{{""-""}}}{{{1/27}}}

Get an LCD of 27

{{{(1*3)/(9*3)}}}{{{""-""}}}{{{1/27}}}

{{{3/27}}}{{{""-""}}}{{{1/27}}}

{{{2/27}}}

Edwin</pre>