What is expand (e^x-e^{-x})^4 ?

The solution to expand (e^x-e^{-x})^4 is e^{4x}-4e^{2x}+6-4e^{-2x}+e^{-4x}

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expand (e^x-e^{-x})^4

expand

(e^{x} - e^{- x})^{4}

e^{4 x} - 4 e^{2 x} + 6 - 4 e^{- 2 x} + e^{- 4 x}

Solution steps

(e^{x} - e^{- x})^{4}

Apply binomial theorem:

(a + b)^{n} = i = 0 \sum n (i n) a^{(n - i)} b^{i}

a = e^{x}, b = - e^{- x}

= i = 0 \sum 4 (i 4) (e^{x})^{(4 - i)} (- e^{- x})^{i}

Expand summation

= \frac{4 !}{0 ! ( 4 - 0 ) !} (e^{x})^{4} (- e^{- x})^{0} + \frac{4 !}{1 ! ( 4 - 1 ) !} (e^{x})^{3} (- e^{- x})^{1} + \frac{4 !}{2 ! ( 4 - 2 ) !} (e^{x})^{2} (- e^{- x})^{2} + \frac{4 !}{3 ! ( 4 - 3 ) !} (e^{x})^{1} (- e^{- x})^{3} + \frac{4 !}{4 ! ( 4 - 4 ) !} (e^{x})^{0} (- e^{- x})^{4}

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (e^{x})^{4} (- e^{- x})^{0} : e^{4 x}

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} (e^{x})^{3} (- e^{- x})^{1} : - 4 e^{2 x}

\frac{4 !}{2 ! ( 4 - 2 ) !} (e^{x})^{2} (- e^{- x})^{2} = 6

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} (e^{x})^{1} (- e^{- x})^{3} : - 4 e^{- 2 x}

Simplify

\frac{4 !}{4 ! ( 4 - 4 ) !} (e^{x})^{0} (- e^{- x})^{4} : e^{- 4 x}

= e^{4 x} - 4 e^{2 x} + 6 - 4 e^{- 2 x} + e^{- 4 x}

A (- 3, - 2) y B (8, 3)

((1 + 4 x)^{- \frac{1}{2}})^{3}

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(r^{2} - x^{2}) d x

(- cos^{2} (x) - 1)^{2}

What is expand (e^x-e^{-x})^4 ?
The solution to expand (e^x-e^{-x})^4 is e^{4x}-4e^{2x}+6-4e^{-2x}+e^{-4x}