What is expand (3-e^{2x^2+x})^4 ?

The solution to expand (3-e^{2x^2+x})^4 is 81-108e^{2x^2+x}+54e^{4x^2+2x}-12e^{6x^2+3x}+e^{8x^2+4x}

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expand (3-e^{2x^2+x})^4

expand

(3 - e^{2 x^{2} + x})^{4}

81 - 108 e^{2 x^{2} + x} + 54 e^{4 x^{2} + 2 x} - 12 e^{6 x^{2} + 3 x} + e^{8 x^{2} + 4 x}

Solution steps

(3 - e^{2 x^{2} + x})^{4}

Apply binomial theorem:

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

a = 3, b = - e^{2 x^{2} + x}

= i = 0 \sum 4 (i 4) \cdot 3^{(4 - i)} (- e^{2 x^{2} + x})^{i}

Expand summation

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

\frac{4 !}{0 ! ( 4 - 0 ) !} \cdot 3^{4} (- e^{2 x^{2} + x})^{0} = 81

Simplify

\frac{4 !}{1 ! ( 4 - 1 ) !} \cdot 3^{3} (- e^{2 x^{2} + x})^{1} : - 108 e^{2 x^{2} + x}

Simplify

\frac{4 !}{2 ! ( 4 - 2 ) !} \cdot 3^{2} (- e^{2 x^{2} + x})^{2} : 54 e^{2 (2 x^{2} + x)}

Simplify

\frac{4 !}{3 ! ( 4 - 3 ) !} \cdot 3^{1} (- e^{2 x^{2} + x})^{3} : - 12 e^{3 (2 x^{2} + x)}

Simplify

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

= 81 - 108 e^{2 x^{2} + x} + 54 e^{2 (2 x^{2} + x)} - 12 e^{3 (2 x^{2} + x)} + e^{4 (2 x^{2} + x)}

Simplify

81 - 108 e^{2 x^{2} + x} + 54 e^{2 (2 x^{2} + x)} - 12 e^{3 (2 x^{2} + x)} + e^{4 (2 x^{2} + x)} : 81 - 108 e^{2 x^{2} + x} + 54 e^{4 x^{2} + 2 x} - 12 e^{6 x^{2} + 3 x} + e^{8 x^{2} + 4 x}

= 81 - 108 e^{2 x^{2} + x} + 54 e^{4 x^{2} + 2 x} - 12 e^{6 x^{2} + 3 x} + e^{8 x^{2} + 4 x}

\frac{2}{3} (a + 3) (b + (- \frac{4}{3}))

\frac{1}{81} (2 + s)^{4}

e^{(y + 1)^{2}} ((y^{2} + y) d x - x d y)

(63 - 5) (25 + 42)

- \frac{a ^{2} ( - 5 a ^{2} b ^{4} - 6 a c + 7 c ^{3} )}{b ^{2} c}

What is expand (3-e^{2x^2+x})^4 ?
The solution to expand (3-e^{2x^2+x})^4 is 81-108e^{2x^2+x}+54e^{4x^2+2x}-12e^{6x^2+3x}+e^{8x^2+4x}