Let's try to simplify that a bit. Rewriting it in its current form it looks like this:
[tex] \frac{3x^3y^{-2}}{5x^{-2}y^3} [/tex].
We could split it up and deal with one problem at a time if we do this:
[tex] \frac{3}{5}*\frac{x^3}{x^{-2}}*\frac{y^{-2}}{y^3} [/tex].
The rule with like bases and division of the exponents is that we subtract the lower exponent from the upper so that would be
[tex] x^{3-(-2)}*y^{-2-3} [/tex]
which gives us new exponents of
[tex] x^5*y^{-5} [/tex].
We rewrite to make that negative exponent a positive by putting it under a 1, so the whole problem with no negatives looks like this now:
[tex] \frac{3x^5}{5y^5} [/tex].
If x = 2 and y = 3, we sub in accordingly:
[tex] \frac{3(2)^5}{5(3)^5} =\frac{3(32)}{5(243)} [/tex]
which simplifies to [tex] \frac{96}{1215}=\frac{32}{405} [/tex], the first choice above.
