Решите, пожалуйста, срочно! Даю 25 баллов!!

Решите задачу:

${(15a {b}^{2} )}^{2}$
${(3c {d}^{2}) }^{3}$
${(4 {a}^{4}b) }^{2}$

${(12 {x}^{4}y) }^{2}$
${(9m {n}^{4} )}^{2}$
${(0.2 \times 5)}^{9} = {1}^{9} = 1$
$( \frac{3}{7} \times \frac{7}{3})^{21} = {1}^{21}$
$\frac{ {7}^{6} \times {3}^{6} }{ {21}^{5} } = \frac{ {21}^{6} }{ {21}^{5} } = 21$

${2}^{ - 2 + ( - 2)} = {2}^{ - 4}$
${( \frac{1}{7} \times 7)}^{ - 2} = {1}^{ - 2} = 1$
${2 }^{ - 3 + ( - 7) - ( - 5) - ( - 4)} = {2}^{ - 1} = 0.5$
${( - 4)}^{ - ( - 2)} \times {2}^{ - 3} = {4}^{2} \times {4}^{ - \frac{3}{2} } = {4}^{ \frac{1}{2} } = 2$
$\frac{ {3}^{ - 5} \times {3}^{ - 2} }{ {3}^{ - 8} } = {3}^{ - 5 + ( - 2) - ( - 8)} = {3}^{1} = 3$
${3}^{ - 2 + 3} = {3}^{1} = 3$
${5}^{ - ( - 2)} \times {5}^{3} \times {5}^{ - 4} = {5}^{2 + 3 - 4} = {5}^{1} = 5$
$(0.2) = {5}^{ - 1} \\ {(0.2)}^{ - 3} = {5}^{ - ( - 3)} = {5}^{3} = 125$
$\frac{ {7}^{ - 3 \times ( - 2)} \times {7}^{ - 5} }{ {7}^{ - 2 + 3} } = \frac{ {7}^{6 - 5} }{7} = \frac{7}{7} = 1$
${(5)}^{ - ( - 2)} \times {5}^{ - 3} \times {5}^{ - 2} = {5}^{2 - 3 - 2} = {5}^{ - 3} = 0.008$