Binomial Expansion of (xy)^7
The binomial theorem provides a formula for expanding expressions of the form (x + y)^n, where n is a nonnegative integer. We can apply this theorem to expand (xy)^7.
The Binomial Theorem
The binomial theorem states:
(x + y)^n = ∑_(k=0)^n (n choose k) * x^(nk) * y^k
where:
 (n choose k) is the binomial coefficient, calculated as n! / (k! * (nk)!).
 ∑_(k=0)^n represents the summation from k = 0 to k = n.
Expanding (x  y)^7

Set up the expansion:
We'll use the binomial theorem with n = 7. We need to find the terms of the form (7 choose k) * x^(7k) * (y)^k for each value of k from 0 to 7.

Calculate the binomial coefficients:
 (7 choose 0) = 1
 (7 choose 1) = 7
 (7 choose 2) = 21
 (7 choose 3) = 35
 (7 choose 4) = 35
 (7 choose 5) = 21
 (7 choose 6) = 7
 (7 choose 7) = 1

Write out the terms:
 k = 0: (7 choose 0) * x^(70) * (y)^0 = x^7
 k = 1: (7 choose 1) * x^(71) * (y)^1 = 7x^6y
 k = 2: (7 choose 2) * x^(72) * (y)^2 = 21x^5y^2
 k = 3: (7 choose 3) * x^(73) * (y)^3 = 35x^4y^3
 k = 4: (7 choose 4) * x^(74) * (y)^4 = 35x^3y^4
 k = 5: (7 choose 5) * x^(75) * (y)^5 = 21x^2y^5
 k = 6: (7 choose 6) * x^(76) * (y)^6 = 7xy^6
 k = 7: (7 choose 7) * x^(77) * (y)^7 = y^7

Combine the terms:
(x  y)^7 = x^7  7x^6y + 21x^5y^2  35x^4y^3 + 35x^3y^4  21x^2y^5 + 7xy^6  y^7
Key Observations
 The coefficients alternate in sign due to the (y) term.
 The exponents of x decrease from 7 to 0, while the exponents of y increase from 0 to 7.
 The sum of the exponents in each term always equals 7.
The expanded form of (xy)^7 helps us understand how the terms combine and how the powers of x and y interact. This expansion is useful in various areas of mathematics, including algebra, calculus, and probability.