${\displaystyle {(a^{n})}^{m}=a^{n\cdot m}\qquad a^{1 \over m}={\sqrt[{m}]{a}}}$

A

${\displaystyle {\left(b^{3 \over 7}\right)}^{28 \over 9}}$${\displaystyle \quad =b^{{\frac {3}{7}}\cdot {\frac {28}{9}}}=b^{\frac {4}{3}}\left(={\sqrt[{3}]{b^{4}}}\right)}$

${\displaystyle \quad {(c^{-5})}^{3}}$${\displaystyle \quad =a^{(-5)\cdot 3}=c^{-15}\qquad {(s^{-5})}^{-4}=s^{(-5)\cdot (-4)}=s^{20}}$

${\displaystyle \quad {(a^{7})}^{3}}$${\displaystyle \quad =a^{7\cdot 3}=a^{21}\qquad {(x^{b})}^{4}=x^{b\ \cdot \ 4}=x^{4b}\qquad {(x^{x})}^{4}=x^{x\ \cdot \ 4}=x^{4x}}$

B

${\displaystyle {\left({\sqrt[{35}]{w^{-5}}}\right)}^{-14}}$${\displaystyle \quad ={\left[{(w^{-5})}^{1 \over 35}\right]}^{-14}=w^{(-5)\cdot {1 \over 35}\cdot {(-14)}}=w^{70 \over 35}=w^{2}}$

${\displaystyle {\sqrt[{4}]{y^{-5}}}^{\ 3}}$${\displaystyle \quad ={\left[{(y^{-5})}^{1 \over 4}\right]}^{3}=y^{(-5)\cdot {1 \over 4}\cdot 3}=y^{-{15 \over 4}}\left(=y^{3{,}75}\right)}$

C

${\displaystyle \left(\left(b^{3 \over 5}\right)^{3}\cdot {\sqrt[{5}]{b^{6}}}\cdot b^{-2}\right)^{29}}$${\displaystyle \quad =\left(b^{{\frac {3}{5}}\cdot 3}\cdot {b^{6 \over 5}}\cdot b^{-2}\right)^{29}=\left(b^{{\frac {9}{5}}+{\frac {6}{5}}-2}\right)^{29}=\dots }$

${\displaystyle \dots =\left(b^{{\cancelto {3}{\ \ {\frac {15}{5}}\ \ }}-2}\right)^{29}=\left(b^{1}\right)^{29}=b^{29}}$

D

${\displaystyle {\dfrac {\sqrt[{3}]{{\Bigl (}b^{15 \over 7}{\Bigr )}^{28}}}{b^{3} \over b^{-8}}}}$${\displaystyle \quad ={\frac {\left(b^{\frac {15}{7}}\right)^{28 \over 3}}{b^{3-(-8)}}}={\frac {b^{{\frac {15}{7}}\cdot {\frac {28}{3}}}}{b^{3+8}}}={\frac {b^{20}}{b^{11}}}=b^{9}}$