What is (sqrt(ab^{-2)})^4 ?

The solution to (sqrt(ab^{-2)})^4 is (a^2)/(b^4)

Popular >

simplify (sqrt(ab^{-2)})^4

simplify

(a b^{- 2})^{4}

\frac{a ^{2}}{b ^{4}}

Solution steps

(a b^{- 2})^{4}

Rewrite

4

2 \cdot 2

= (a b^{- 2})^{2 \cdot 2}

Apply exponent rule:

a^{b c} = (a^{b})^{c},

assuming

a \geq 0

(a b^{- 2})^{2 \cdot 2} = ((a b^{- 2})^{2})^{2}

= ((a b^{- 2})^{2})^{2}

Apply exponent rule:

assuming

n i s o dd or a \geq 0

(a b^{- 2})^{2} = a b^{- 2}

= (a b^{- 2})^{2}

Apply exponent rule:

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

(a b^{- 2})^{2} = a^{2} (b^{- 2})^{2}

= a^{2} (b^{- 2})^{2}

Simplify

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

= a^{2} \frac{1}{b ^{4}}

Simplify

a^{2} \frac{1}{b ^{4}} : \frac{a ^{2}}{b ^{4}}

= \frac{a ^{2}}{b ^{4}}

(2 x^{2} yz) (3 x y z^{2})

(4 m^{8}) (16 n^{8})

(3 - x)^{- 2}

(x - 2) (x - 2)

\frac{( - 9 ) ^{5}}{9 ^{- 5}} \cdot ((- 9)^{3})^{- 3}