Answer
$$\frac{1}{12x^{10}y^{3}}$$
Work Step by Step
$$\frac{(3x^{3}y^{2})^{-1}(2x^{3}y)^{-2}}{(xy^{2})^{-5}(x^{2}y^{3})^{3}}$$
Simplify the numerator: $(3x^{3}y^{2})^{-1}(2x^{3}y)^{-2}$
Recall the power rule: $(a^{m})^{n}=a^{mn}$
Thus,
$$(3x^{3}y^{2})^{-1}(2x^{3}y)^{-2} = (3^{-1}x^{3(-1)}y^{2(-1)})(2^{-2}x^{3(-2)}y^{-2}) = (3^{-1}x^{-3}y^{-2})(2^{-2}x^{-6}y^{-2})$$
Recall the product rule: $a^{m}⋅a^{n}=a^{m+n}$
Thus,
$$(3^{-1}x^{-3}y^{-2})(2^{-2}x^{-6}y^{-2}) = (3^{-1})(2^{-2})(x^{-3-6})(y^{-2-2}) = (3^{-1})(2^{-2})(x^{-9})(y^{-4})$$
Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$
Thus,
$$(3^{-1})(2^{-2})(x^{-9})(y^{-4}) = \frac{1}{(3^{1})(2^{2})(x^{9})(y^{4})}$$
Simplify the denominator:$(xy^{2})^{-5}(x^{2}y^{3})^{3}$
Using power rule: $(x^{-5}y^{2(-5)})(x^{2(3)}y^{3(3)})=(x^{-5}y^{-10})(x^{6}y^{9})$
Using the product rule: $(x^{-5}y^{-10})(x^{6}y^{9}) = x^{-5+6}y^{-10+9} = x^{1}y^{-1}$
Rewrite the equation using the simplified forms.
$$(\frac{1}{(3^{1})(2^{2})(x^{9})(y^{4})})(x^{1}y^{-1}) = \frac{x^{1}y^{-1}}{(3^{1})(2^{2})(x^{9})(y^{4})} = \frac{x^{1}y^{-1}}{12(x^{9})(y^{4})}$$
Recall the quotient rule: $\frac{a^{m}}{a^{n}}=a^{m-n}$ and $\frac{a^{n}}{a^{m}}=\frac{1}{a^{m+n}}$ if $m>n$
Thus,
$$=\frac{x^{1}y^{-1}}{12(x^{9})(y^{4})} = \frac{1}{12(x^{9+1})(y^{4-1})} = \frac{1}{12x^{10}y^{3}}$$
