(x - 2y)(x - 2y)(x - 2y)
but it's better to use the binomial expansion theorem. For this the binomial expansion looks like this
C(3,0)(x)³(-2y)⁰ + C(3,1)(x)²(-2y)¹ + C(3,2)(x)¹(-2y)² + C(3,3)(x)⁰(-2y)³
C(n,k) = n! / k!(n - k)!
This is called the binomial coefficient but it also shows up in combinatorics as combinations of groups.
(1)(x³)(1) + (3)(x²)(-2y) + (3)(x)(4y²) + (1)(1)(-8y³)
x³ - 6x² + 12xy² - 8y³
It looks complicated, but trust me the first idea is much worse when you get to higher powers whereas the second doesn't get any more complicated, you just add more terms.
