What is [(1/2 x^{-2})^3(4xy^{-1})^2]^2 ?

The solution to [(1/2 x^{-2})^3(4xy^{-1})^2]^2 is 4/(x^8y^4)

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simplify [(1/2 x^{-2})^3(4xy^{-1})^2]^2

simplify

[(\frac{1}{2} x^{- 2})^{3} (4 x y^{- 1})^{2}]^{2}

\frac{4}{x ^{8} y ^{4}}

Solution steps

((\frac{1}{2} x^{- 2})^{3} (4 x y^{- 1})^{2})^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

((\frac{1}{2} x^{- 2})^{3} (4 x y^{- 1})^{2})^{2} = ((\frac{1}{2} x^{- 2})^{3})^{2} ((4 x y^{- 1})^{2})^{2}

= ((\frac{1}{2} x^{- 2})^{3})^{2} ((4 x y^{- 1})^{2})^{2}

Simplify

((\frac{1}{2} x^{- 2})^{3})^{2} : \frac{1}{64 x ^{12}}

= \frac{1}{64 x ^{12}} ((4 x y^{- 1})^{2})^{2}

Simplify

((4 x y^{- 1})^{2})^{2} : \frac{256 x ^{4}}{y ^{4}}

= \frac{1}{64 x ^{12}} \cdot \frac{256 x ^{4}}{y ^{4}}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{1 \cdot 256 x ^{4}}{64 x ^{12} y ^{4}}

Multiply the numbers:

1 \cdot 256 = 256

= \frac{256 x ^{4}}{64 x ^{12} y ^{4}}

Divide the numbers:

\frac{256}{64} = 4

= \frac{4 x ^{4}}{x ^{12} y ^{4}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = \frac{1}{x ^{b - a}}

\frac{x ^{4}}{x ^{12}} = \frac{1}{x ^{12 - 4}}

= \frac{4}{y ^{4} x ^{12 - 4}}

Subtract the numbers:

12 - 4 = 8

= \frac{4}{x ^{8} y ^{4}}

\frac{2}{3} \cdot \frac{3}{4} \cdot \frac{5}{6}

(6 a - 5 b) (6 a + 5 b)

\frac{6 ^{9}}{6 ^{3}}

5 x^{4} - 20 x^{3}

1 5^{2} + 1 2^{2}

What is [(1/2 x^{-2})^3(4xy^{-1})^2]^2 ?
The solution to [(1/2 x^{-2})^3(4xy^{-1})^2]^2 is 4/(x^8y^4)