What is expand (e^{xy}+e^y)^4 ?

The solution to expand (e^{xy}+e^y)^4 is e^{4xy}+4e^{3xy+y}+6e^{2xy+2y}+4e^{xy+3y}+e^{4y}

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expand (e^{xy}+e^y)^4

expand

(e^{x y} + e^{y})^{4}

e^{4 x y} + 4 e^{3 x y + y} + 6 e^{2 x y + 2 y} + 4 e^{x y + 3 y} + e^{4 y}

Solution steps

(e^{x y} + e^{y})^{4}

Apply binomial theorem:

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

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

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

Expand summation

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

Simplify

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

= e^{4 x y} + 4 e^{3 x y + y} + 6 e^{2 (x y + y)} + 4 e^{x y + 3 y} + e^{4 y}

e^{2 (x y + y)} = e^{2 x y + 2 y}

= e^{4 x y} + 4 e^{3 x y + y} + 6 e^{2 x y + 2 y} + 4 e^{x y + 3 y} + e^{4 y}

\frac{1}{12} e^{4 x} (2 x - 1)

(x ln (x) + x y) d y

\frac{1}{x ( 10 + s )}

(e^{x} + y) d x

(2 a + 3) (- 2 - a = 5 a^{2})

What is expand (e^{xy}+e^y)^4 ?
The solution to expand (e^{xy}+e^y)^4 is e^{4xy}+4e^{3xy+y}+6e^{2xy+2y}+4e^{xy+3y}+e^{4y}