What is (a^{-7}+b)^{-4} ?

The solution to (a^{-7}+b)^{-4} is (a^{28})/((a^7b+1)^4)

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simplify (a^{-7}+b)^{-4}

simplify

(a^{- 7} + b)^{- 4}

\frac{a ^{28}}{( a ^{7} b + 1 ) ^{4}}

Solution steps

(a^{- 7} + b)^{- 4}

Apply exponent rule:

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

(a^{- 7} + b)^{- 4} = \frac{1}{( a ^{- 7} + b ) ^{4}}

= \frac{1}{( a ^{- 7} + b ) ^{4}}

Simplify

(a^{- 7} + b)^{4} : \frac{( b a ^{7} + 1 ) ^{4}}{a ^{28}}

= \frac{1}{\frac{( b a ^{7} + 1 ) ^{4}}{a ^{28}}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{a ^{28}}{( a ^{7} b + 1 ) ^{4}}

b^{2} - 6 b

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

\frac{d ( 3 2 )}{( 3 + 6 )}

(- x y + 0.5)^{2}

(\frac{50 x ^{4}}{25 \cdot x ^{2}}) - 12 x^{2}

What is (a^{-7}+b)^{-4} ?
The solution to (a^{-7}+b)^{-4} is (a^{28})/((a^7b+1)^4)