What is (1-a^{m-n})^{-1}+(1-a^{n-m}) ?

The solution to (1-a^{m-n})^{-1}+(1-a^{n-m}) is 1/(1-a^{m-n)}+1-a^{n-m}

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simplify (1-a^{m-n})^{-1}+(1-a^{n-m})

simplify

(1 - a^{m - n})^{- 1} + (1 - a^{n - m})

\frac{1}{1 - a ^{m - n}} + 1 - a^{n - m}

Solution steps

(1 - a^{m - n})^{- 1} + (1 - a^{n - m})

Apply rule:

a + (b + c) = a + b + c

(1 - a^{m - n})^{- 1} + (1 - a^{n - m}) = (1 - a^{m - n})^{- 1} + 1 - a^{n - m}

= (1 - a^{m - n})^{- 1} + 1 - a^{n - m}

Apply exponent rule:

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

(1 - a^{m - n})^{- 1} = \frac{1}{1 - a ^{m - n}}

= \frac{1}{1 - a ^{m - n}} + 1 - a^{n - m}

(3 w + 5)^{3}

2 (\frac{a ^{2} + 1}{( a ^{2} )}) - (\frac{a - 1}{a}) - 7

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

\frac{( n ^{2} - 5 n - 14 )}{( n ^{2} - 9 n + 14 )}

35 y^{2} + 13 y - 12

What is (1-a^{m-n})^{-1}+(1-a^{n-m}) ?
The solution to (1-a^{m-n})^{-1}+(1-a^{n-m}) is 1/(1-a^{m-n)}+1-a^{n-m}