What is expand (e^{x^2}-1)^4 ?

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

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expand (e^{x^2}-1)^4

expand

(e^{x^{2}} - 1)^{4}

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

Solution steps

(e^{x^{2}} - 1)^{4}

Apply binomial theorem:

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

a = e^{x^{2}}, b = - 1

= i = 0 \sum 4 (i 4) (e^{x^{2}})^{(4 - i)} (- 1)^{i}

Expand summation

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

Simplify

\frac{4 !}{0 ! ( 4 - 0 ) !} (e^{x^{2}})^{4} (- 1)^{0} : e^{4 x^{2}}

Simplify

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

Simplify

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

Simplify

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

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

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

\frac{32 ( 2 n + 1 ) ( n + 1 )}{3 n ^{2}}

(sin (x) + x cos (x)) (x + cos (x))

P_{15} (P_{6} + P_{7} x + x^{2})^{3}

(tan (x) - 4) (tan (x) + 4) (tan (x) - 2) (tan (x) + 2)

5 x (x - 8 x)

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