Expand the following
4(2a7 - 6b4)3
Since 4(2a7 - 6b4)3 is a binomial
Use the binomial theorem to expand this.
a(x + y)n = aΣ[k = 0 to n]C(n,k) xn-kyk
where
C(n,k) = | n! |
k!(n - k)! |
n = 3, x = 2a7, a = 4, and y = -6b4.
Expanding terms, we get:
4C(3,0)x3-0y0 = 4C(3,0)x3y0
C(3,0) = | 3! |
0!(3 - 0)! |
C(3,0) = 1
Click to see C(3,0)
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)
4C(3,0)x3y0 = (4 * 1 * 8 * 1)(1)
Anything raised to a 0 power = 1
4C(3,0)x3y0 = 32a21
4C(3,1)x3-1y1 = 4C(3,1)x2y1
C(3,1) = | 3! |
1!(3 - 1)! |
C(3,1) = 3
Click to see C(3,1)
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)
4C(3,1)x2y1 = (4 * 3 * 4 * -6)(a14b4)
4C(3,1)x2y1 = -288a14b4
4C(3,2)x3-2y2 = 4C(3,2)x1y2
C(3,2) = | 3! |
2!(3 - 2)! |
C(3,2) = 3
Click to see C(3,2)
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)
4C(3,2)x1y2 = (4 * 3 * 2 * 36)(a7b8)
4C(3,2)x1y2 = 864a7b8
4C(3,3)x3-3y3 = 4C(3,3)x0y3
C(3,3) = | 3! |
3!(3 - 3)! |
C(3,3) = 1
Click to see C(3,3)
4C(3,3)x0y3 = (4)C(3,3)(2a7)3 - 3(-6b4)3
4C(3,3)x0y3 = (4)(1)(2a7)0(-6b4)3
4C(3,3)x0y3 = (4)(1)(1)(-216b12)
4C(3,3)x0y3 = (4 * 1 * 1 * -216)(b12)
4C(3,3)x0y3 = -864b12
4(2a7 - 6b4)3 = 32a21 - 288a14b4 + 864a7b8 - 864b12
************ End Binomial Expansion ********************
32a21 - 288a14b4 + 864a7b8 - 864b12
32a21 - 288a14b4 + 864a7b8 - 864b12
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