Answer
(a) $\dfrac{x^6}{y^{12}}$
(b) $\dfrac{x^9}{8y^{14}}$
(c) $\dfrac{1}{32ab^8}$
Work Step by Step
RECALL:
(1) $(ab)^m=a^mb^m$
(2) $(a^m)^n=a^{mn}$
(3) $a^{-m} = \dfrac{1}{a^m}$
(4) $\dfrac{a^m}{a^n} = a^{m-n}, a\ne0$
(5) $a^m\cdot a^n=a^{m+n}$
(a) Use rule (1) above to obtain:
$=(x^{-2})^{-3}(y^4)^{-3}$
Use rule (2) above to obtain:
$=x^{-2(-3)}y^{4(-3)}
\\=x^{6}y^{-12}$
Use rule (3) above to obtain:
$=x^{6} \cdot \dfrac{1}{y^{12}}
\\=\dfrac{x^6}{y^{12}}$
(b) use rule (1) above to obtain:
$=(y^2)^{-1}(2^{-3})(x^{-3})^{-3}(y^4)^{-3}$
Use rule (2) above to obtain:
$=y^{2(-1)}(2^{-3})(x^{-3(-3)})y^{4(-3)}
\\=y^{-2}(2^{-3})x^{9}y^{-12}$
Use rule (5) above to obtain:
$\\=y^{-2}(2^{-3})x^{9}y^{-12}
\\=(2^{-3})x^9y^{-2+(-12)}
\\=(2^{-3})x^9y^{-14}$
Use rule (3) above to obtain:
$=\dfrac{1}{2^3} \cdot x^9 \cdot \dfrac{1}{y^{14}}
\\=\dfrac{1}{8} \cdot x^9 \cdot \dfrac{1}{y^{14}}
\\=\dfrac{x^9}{8y^{14}}$
(c) Use rule (1) above to obtain:
$=\left(\dfrac{2^{-3} (a^{-1})^{-3}}{(b^{-2})^{-3}}\right) \left(\dfrac{(b^{-1})^2}{2^2(a^2)^2}\right)$
Use rule (2) above to obtain:
$=\left(\dfrac{2^{-3} (a^{-1(-3)}}{(b^{-2(-3)})}\right) \left(\dfrac{(b^{-1(2)})}{4(a^{2(2)}}\right)
\\=\left(\dfrac{2^{-3} a^{3}}{b^{6}}\right) \left(\dfrac{b^{-2}}{4a^{4}}\right)
\\=\dfrac{2^{-3}a^{3}b^{-2}}{4a^4b^6}$
Use rule (4) above to obtain:
$=\dfrac{2^{-3}}{4}a^{3-4}b^{-2-6}
\\=\dfrac{2^{-3}}{4}a^{-1}b^{-8}$
Use rule (3) above to obtain:
$=\dfrac{1}{2^3(4)} \cdot \dfrac{1}{a} \cdot \dfrac{1}{b^{8}}
\\=\dfrac{1}{8(4)ab^8}
\\=\dfrac{1}{32ab^8}$
