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Expand 4(2a7 - 6b4)3

Since 4(2a7 - 6b4)3 is a binomial expansion, we can use the binomial theorem to expand this expression.
The formula for this is a(x + y)n = aΣ[k = 0 to n]C(n,k) xn-kyk where
 C(n,k) = n! k!(n - k)!

In this case, n = 3, x = 2a7, a = 4, and y = -6b4. Expanding terms, we get:

k = 0 expansion:
4C(3,0)x3-0y0 = 4C(3,0)x3y0

With n = 3 and k = 0, C(3,0) becomes:
 C(3,0) = 3! 0!(3 - 0)!

Simplifying the rest of our expansion, we get:
4C(3,0)x3y0 = (4)C(3,0)(2a7)3 - 0(-6b4)0
4C(3,0)x3y0 = (4)(1)(2a7)3(-6b4)0
4C(3,0)x3y0 = (4)(1)(8a21)(1)

Group constants and powers from all terms
4C(3,0)x3y0 = (4 * 1 * 8 * 1)(1)← Anything raised to a 0 power = 1
4C(3,0)x3y0 = 32a21

k = 1 expansion:
4C(3,1)x3-1y1 = 4C(3,1)x2y1

With n = 3 and k = 1, C(3,1) becomes:
 C(3,1) = 3! 1!(3 - 1)!

Simplifying the rest of our expansion, we get:
4C(3,1)x2y1 = (4)C(3,1)(2a7)3 - 1(-6b4)1
4C(3,1)x2y1 = (4)(3)(2a7)2(-6b4)1
4C(3,1)x2y1 = (4)(3)(4a14)(-6b4)

Group constants and powers from all terms
4C(3,1)x2y1 = (4 * 3 * 4 * -6)(a14b4)
4C(3,1)x2y1 = -288a14b4

k = 2 expansion:
4C(3,2)x3-2y2 = 4C(3,2)x1y2

With n = 3 and k = 2, C(3,2) becomes:
 C(3,2) = 3! 2!(3 - 2)!

Simplifying the rest of our expansion, we get:
4C(3,2)x1y2 = (4)C(3,2)(2a7)3 - 2(-6b4)2
4C(3,2)x1y2 = (4)(3)(2a7)1(-6b4)2
4C(3,2)x1y2 = (4)(3)(2a7)(36b8)

Group constants and powers from all terms
4C(3,2)x1y2 = (4 * 3 * 2 * 36)(a7b8)
4C(3,2)x1y2 = 864a7b8

k = 3 expansion:
4C(3,3)x3-3y3 = 4C(3,3)x0y3

With n = 3 and k = 3, C(3,3) becomes:
 C(3,3) = 3! 3!(3 - 3)!