What is (ab^{-1}+ba^{-1})(a^{-2}+b^{-2})^{-1} ?

The solution to (ab^{-1}+ba^{-1})(a^{-2}+b^{-2})^{-1} is ab

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simplify (ab^{-1}+ba^{-1})(a^{-2}+b^{-2})^{-1}

simplify

(a b^{- 1} + b a^{- 1}) (a^{- 2} + b^{- 2})^{- 1}

ab

Solution steps

(a b^{- 1} + b a^{- 1}) (a^{- 2} + b^{- 2})^{- 1}

Simplify

a b^{- 1} + b a^{- 1} : \frac{a}{b} + \frac{b}{a}

= (\frac{a}{b} + \frac{b}{a}) (a^{- 2} + b^{- 2})^{- 1}

Simplify

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

= (\frac{a}{b} + \frac{b}{a}) \frac{a ^{2} b ^{2}}{b ^{2} + a ^{2}}

Apply the fraction rule:

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

= \frac{( \frac{a}{b} + \frac{b}{a} ) a ^{2} b ^{2}}{b ^{2} + a ^{2}}

Simplify

(\frac{a}{b} + \frac{b}{a}) a^{2} b^{2} : (a^{2} + b^{2}) ab

= \frac{( a ^{2} + b ^{2} ) ab}{b ^{2} + a ^{2}}

Apply the commutative law:

a^{2} + b^{2} = b^{2} + a^{2}

= \frac{( b ^{2} + a ^{2} ) ab}{b ^{2} + a ^{2}}

Cancel the common factor:

b^{2} + a^{2}

= ab

\frac{2 y}{3} + \frac{2 y}{5} < \frac{16 y}{15} - \frac{1}{3}

24 \geq - 4 n

f = - \frac{1}{2} (2)^{2}

\frac{3 x - 5}{2} = \frac{2 x + 7}{4}

- 5 = - 5 x

What is (ab^{-1}+ba^{-1})(a^{-2}+b^{-2})^{-1} ?
The solution to (ab^{-1}+ba^{-1})(a^{-2}+b^{-2})^{-1} is ab