What is (sqrt((1-a)/((a+1)^{-1))})^4 ?

The solution to (sqrt((1-a)/((a+1)^{-1))})^4 is (1-a)^2(a+1)^2

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simplify (sqrt((1-a)/((a+1)^{-1))})^4

simplify

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

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

Solution steps

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

Rewrite

4

2 \cdot 2

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

Apply exponent rule:

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

assuming

a \geq 0

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

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

Apply exponent rule:

assuming

n i s o dd or a \geq 0

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Simplify

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

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

Simplify

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

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

(a b^{2}) (a^{2} b) (ab)

(100 \cdot 1 0^{- 12}) \cdot (- 6 e^{- t})

75 (0.86)^{- 3}

2.52 x 1 0^{- 5}

What is (sqrt((1-a)/((a+1)^{-1))})^4 ?
The solution to (sqrt((1-a)/((a+1)^{-1))})^4 is (1-a)^2(a+1)^2