Answer
$$(\frac{x^{3-n}}{x^{6-n}})^{-2}=x^{6}$$
Work Step by Step
$$(\frac{x^{3-n}}{x^{6-n}})^{-2}$$
Recall the quotient rule: $\frac{a^{m}}{a^{n}}=a^{m-n}$ and $\frac{a^{n}}{a^{m}}=\frac{1}{a^{m-n}}$
Thus,
$$(\frac{x^{3-n}}{x^{6-n}})^{-2}$$ $$=(\frac{1}{x^{(6-n)-(3-n)}})^{-2}$$ $$=(\frac{1}{x^{6-n-3+n}})^{-2}$$ $$=(\frac{1}{x^{3}})^{-2}$$
Recall the negative exponent rule: $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}} = a^{n}$
Thus,
$$=(\frac{1}{x^{3}})^{-2}$$ $$=\frac{1}{(\frac{1}{x^{3}})^{2}}$$ $$=(x^{3})^{2}$$
Recall the power rule: $(a^{m})^{n}=a^{mn}$
Thus,
$$=(x^{3})^{2}$$ $$=x^{6}$$
