Question

prove that [(1/2)^2]^3 × (1/3)^-4 × 3^-2 × 1/6 = 3/128​

Answer 1

Step-by-step explanation:

$\huge\bf\underbrace{\overbrace{\pink{❄A} \orange{N}\blue{S}\red{W}\green{E}\purple{R❄}}} \\ \\ { { ((\frac{1}{2}) }^{2}) }^{3} \times { (\frac{1}{3}) }^{ - 4} \times {3}^{ - 2} \times \frac{1}{6} = \frac{3}{128} \\ = \: \: \: \: \frac{ {(\frac{1}{2})}^{6} \times \frac{1}{6} }{ {(\frac{1}{3})}^{4} \times {3}^{2} } = \frac{3}{128} \\ = \: \: \: \: \frac{ \frac{1}{64} \times \frac{1}{6} }{ \frac{1}{81} \times 9 } = \frac{3}{128} \\ = \: \: \: \: \frac{81 \times \frac{1}{9} }{64 \times 6} = \frac{3}{128} \\ = \: \: \: \: \frac{9}{64 \times 6} = \frac{3}{128} \\ = \: \: \: \: \frac{3}{64 \times 2} = \frac{3}{128} \\ = \: \: \: \: \boxed{ \bold{ \underline{ \underline{ \frac{3}{128} = \frac{3}{128} }}}}$