What is expand (e^x-1)^4 ?

The solution to expand (e^x-1)^4 is e^{4x}-4e^{3x}+6e^{2x}-4e^x+1

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expand (e^x-1)^4

expand

(e^{x} - 1)^{4}

e^{4 x} - 4 e^{3 x} + 6 e^{2 x} - 4 e^{x} + 1

Solution steps

(e^{x} - 1)^{4}

Apply binomial theorem:

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

a = e^{x}, b = - 1

= i = 0 \sum 4 (i 4) (e^{x})^{(4 - i)} (- 1)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (e^{x})^{4} (- 1)^{0} : e^{4 x}

Simplify

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

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} (e^{x})^{2} (- 1)^{2} : 6 e^{2 x}

Simplify

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

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

= e^{4 x} - 4 e^{3 x} + 6 e^{2 x} - 4 e^{x} + 1

(4 y^{3} - 8 y^{2} - 3 y + 1) (y + \frac{1}{2})

cos (2 x) (t - π)

x (sin (u) + u cos (u)) + y (cos (u) - u sin (u)) + z

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

3 t (x) x^{^{''}} + 2 (t (x) - 3) x^{^{'}} - 2 x

What is expand (e^x-1)^4 ?
The solution to expand (e^x-1)^4 is e^{4x}-4e^{3x}+6e^{2x}-4e^x+1