Expand the following:
(x-4)^3
(x-4)^3 = sum_(k=0)^3 binomial(3, k) x^(3-k)×(-4)^k = binomial(3, 0) x^3 (-4)^0+binomial(3, 1) x^2 (-4)^1+binomial(3, 2) x^1 (-4)^2+binomial(3, 3) x^0 (-4)^3:
binomial(3, 0) x^3-4 binomial(3, 1) x^2+16 binomial(3, 2) x-64 binomial(3, 3)
binomial(3, 0) = 1, binomial(3, 1) = 3, binomial(3, 2) = 3 and binomial(3, 3) = 1:
x^3-3 4 x^2+3 (-4)^2 x+(-4)^3
(-4)^2 = 16:
x^3-4 3 x^2+16 3 x+(-4)^3
(-4)^3 = (-1)^3×4^3 = -4^3:
x^3-4 3 x^2+16 3 x+-4^3
4^3 = 4×4^2:
x^3-4 3 x^2+16 3 x-4×4^2
4^2 = 16:
x^3-4 3 x^2+16 3 x-16 4
4×16 = 64:
x^3-4 3 x^2+16 3 x-64
16×3 = 48:
x^3-4 3 x^2+48 x-64
-4×3 = -12:
Answer: | x^3+-12 x^2+48 x-64
