Answer
$$\frac{(x^{12})(4y^{8}+1)}{y^{12}}$$
Work Step by Step
$$(\frac{2^{-1}x^{-2}y}{x^{4}y^{-1}})^{-2}+ (\frac{xy^{-3}}{x^{-3}y})^{3}$$
Simplify the first term: $(\frac{2^{-1}x^{-2}y}{x^{4}y^{-1}})^{-2}$
Recall the power rule: $(a^{m})^{n}=a^{mn}$
Thus,
$$(\frac{2^{-1}x^{-2}y}{x^{4}y^{-1}})^{-2} = \frac{2^{-1(-2)}x^{-2(-2)}y^{-2}}{x^{4(-2)}y^{-1(-2)}} = \frac{2^{2}x^{4}y^{-2}}{x^{-8}y^{2}}$$
Simplify the second term: $ (\frac{xy^{-3}}{x^{-3}y})^{3}$
Recall the power rule: $(a^{m})^{n}=a^{mn}$
Thus,
$$ (\frac{xy^{-3}}{x^{-3}y})^{3} = \frac{x^{3}y^{-3(3)}}{x^{-3(3)}y^{3}} = \frac{x^{3}y^{-9}}{x^{-9}y^{3}}$$
Rewrite the equation:
$$\frac{2^{2}x^{4}y^{-2}}{x^{-8}y^{2}} + \frac{x^{3}y^{-9}}{x^{-9}y^{3}}$$
$$=\frac{4x^{4}y^{-2}}{x^{-8}y^{2}} +\frac{x^{3}y^{-9}}{x^{-9}y^{3}}$$
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,
$$=\frac{4x^{4}y^{-2}}{x^{-8}y^{2}} +\frac{x^{3}y^{-9}}{x^{-9}y^{3}}$$
$$=\frac{4x^{4+8}}{y^{2+2}} +\frac{x^{3+9}}{y^{3+9}}$$
$$=\frac{4x^{12}}{y^{4}} +\frac{x^{12}}{y^{12}}$$
Get the LCD through prime factorization:
$y^{4}: y^{2}\cdot y^{2}$
$y^{12}: y^{4}\cdot y^{3} = y^{2}\cdot y^{2}\cdot y^{3}$ --> LCD
Adjust the fractions based on the LCD.
$$=\frac{4x^{12}}{y^{4}} +\frac{x^{12}}{y^{12}}$$
$$=\frac{4x^{12}y^{8}}{y^{12}} +\frac{x^{12}}{y^{12}}$$
Combine the terms.
$$=\frac{(4x^{12}y^{8})+x^{12}}{y^{12}}$$
Factor $x^{12}$.
$$=\frac{(x^{12})(4y^{8}+1)}{y^{12}}$$
