What is (4x^{-2}(yz)^{-1})/(2^3x^4y) ?

The solution to (4x^{-2}(yz)^{-1})/(2^3x^4y) is 1/(2x^6y^2z)

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simplify (4x^{-2}(yz)^{-1})/(2^3x^4y)

simplify

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

\frac{1}{2 x ^{6} y ^{2} z}

Solution steps

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

Factor the number:

4 = 2^{2}

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

\frac{2 ^{2}}{2 ^{3}} = \frac{1}{2}

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

Simplify

\frac{x ^{- 2}}{x ^{4}} : \frac{1}{x ^{6}}

= \frac{( yz ) ^{- 1}}{2 x ^{6} y}

Apply exponent rule:

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

(yz)^{- 1} = y^{- 1} z^{- 1}

= \frac{y ^{- 1} z ^{- 1}}{2 x ^{6} y}

Simplify

\frac{y ^{- 1}}{y} : \frac{1}{y ^{2}}

= \frac{z ^{- 1}}{2 x ^{6} y ^{2}}

Simplify

\frac{z ^{- 1}}{2 x ^{6} y ^{2}} : \frac{1}{z \cdot 2 x ^{6} y ^{2}}

= \frac{1}{z \cdot 2 x ^{6} y ^{2}}

= \frac{1}{2 x ^{6} y ^{2} z}

\frac{x ^{- 5}}{x ^{4}}

2 x^{2} + 2 x + 6 = 0

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

(b - \frac{1}{2})^{4}

12 x - 4 = 8 x + 24

What is (4x^{-2}(yz)^{-1})/(2^3x^4y) ?
The solution to (4x^{-2}(yz)^{-1})/(2^3x^4y) is 1/(2x^6y^2z)