The binomial theorem states that
#(a+b)^n = sum_(i=0)^n C(n, i) a^ib^(n-i) #
where #C(n,i)# is #n# choose #i#.
Let's right out the coefficients of the #C(5, i)#. I do this based on my memory of Pascal's triangle. 1, 5, 10, 10, 5, 1
From this, we can write out each of the terms, with #a = x# and #b = -5y# and #n = 5#:
#(x - 5y)^5 = sum_(i=0)^5 C(5, i) x^i (-5y)^(5-i) #
# = 1 x^0 (-5y)^5 + 5 x^1 (-5y)^4 + 10 x^2 (-5y)^3 + 10 x^3 (-5y)^2 + 5 x^4 (-5y)^1 + 1 x^5 (-5y)^0 #
# = -5^5y^5 + 5^5xy^4 - 2 * 5^4 x^2 y^3 + 2 * 5^3 x^3 y^2 - 5^2x^4 y + x^5#
# = x^5 - 25x^4 y + 250 x^3y^2 - 1250 x^2 y^3 + 3125xy^4 - 3125 y^5 #
