Answer
$$\frac{1}{128x^{7}y^{16}}$$
Work Step by Step
$$\frac{(2x^{2}y^{4})^{-1}(4xy^{3})^{-3}}{(x^{2}y)^{-5}(x^{3}y^{2})^{4}}$$
Simplify the numerator: $(2x^{2}y^{4})^{-1}(4xy^{3})^{-3}$
Recall the power rule: $(a^{m})^{n}=a^{mn}$
Thus,
$$(2x^{2}y^{4})^{-1}(4xy^{3})^{-3} = (2^{-1}x^{2(-1)}y^{4(-1)})(4^{-3}x^{-3}y^{3(-3)}) = (2^{-1}x^{-2}y^{-4})(4^{-3}x^{-3}y^{-9})$$
Recall the product rule: $a^{m}⋅a^{n}=a^{m+n}$
Thus,
$$(2^{-1})(4^{-3})(x^{-2-3})(y^{-4-9}) = (2^{-1})(4^{-3})(x^{-5})(y^{-13})$$
Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$
Thus,
$$(2^{-1})(4^{-3})(x^{-5})(y^{-13}) = \frac{1}{(2^{1})(4^{3})(x^{5})(y^{13})}$$
Simplify the denominator:$(x^{2}y)^{-5}(x^{3}y^{2})^{4}$
Using power rule: $(x^{2(-5)}y^{-5})(x^{3(4)}y^{2(4)}) = (x^{-10}y^{-5})(x^{12}y^{8})$
Using the product rule: $(x^{-10}y^{-5})(x^{12}y^{8}) = x^{-10+12}y^{-5+8} = x^{2}y^{3}$
Rewrite the equation using the simplified forms.
$$(\frac{1}{(2^{1})(4^{3})(x^{5})(y^{13})})(x^{2}y^{3}) = \frac{x^{2}y^{3}}{(2^{1})(4^{3})(x^{5})(y^{13})}$$
$$=\frac{x^{2}y^{3}}{(2)(64)(x^{5})(y^{13})}$$
$$=\frac{x^{2}y^{3}}{128(x^{5})(y^{13})}$$
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,
$$=\frac{x^{2}y^{3}}{128(x^{5})(y^{13})} = \frac{1}{128(x^{5+2})(y^{13+3})} = \frac{1}{128x^{7}y^{16}}$$
