Expand 4(2a

^{7} - 6b

^{4})

^{3} Since 4(2a

^{7} - 6b

^{4})

^{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) x

^{n-k}y

^{k} where

In this case, n = 3, x = 2a

^{7}, a = 4, and y = -6b

^{4}. Expanding terms, we get:

__k = 0 expansion:__4C(3,0)x

^{3-0}y

^{0} = 4C(3,0)x

^{3}y

^{0}__With n = 3 and k = 0, C(3,0) becomes:__C(3,0) = 1

Click here to see Combination Math for C(3,0)__Simplifying the rest of our expansion, we get:__4C(3,0)x

^{3}y

^{0} = (4)C(3,0)(2a

^{7})

^{3 - 0}(-6b

^{4})

^{0}4C(3,0)x

^{3}y

^{0} = (4)(1)(2a

^{7})

^{3}(-6b

^{4})

^{0}4C(3,0)x

^{3}y

^{0} = (4)(1)(8a

^{21})(1)

__Group constants and powers from all terms__4C(3,0)x

^{3}y

^{0} = (4 * 1 * 8 * 1)(1)← Anything raised to a 0 power = 1

4C(3,0)x

^{3}y

^{0} = 32a

^{21}

__k = 1 expansion:__4C(3,1)x

^{3-1}y

^{1} = 4C(3,1)x

^{2}y

^{1}__With n = 3 and k = 1, C(3,1) becomes:__C(3,1) = 3

Click here to see Combination Math for C(3,1)__Simplifying the rest of our expansion, we get:__4C(3,1)x

^{2}y

^{1} = (4)C(3,1)(2a

^{7})

^{3 - 1}(-6b

^{4})

^{1}4C(3,1)x

^{2}y

^{1} = (4)(3)(2a

^{7})

^{2}(-6b

^{4})

^{1}4C(3,1)x

^{2}y

^{1} = (4)(3)(4a

^{14})(-6b

^{4})

__Group constants and powers from all terms__4C(3,1)x

^{2}y

^{1} = (4 * 3 * 4 * -6)(a

^{14}b

^{4})

4C(3,1)x

^{2}y

^{1} = -288a

^{14}b

^{4}

__k = 2 expansion:__4C(3,2)x

^{3-2}y

^{2} = 4C(3,2)x

^{1}y

^{2}__With n = 3 and k = 2, C(3,2) becomes:__C(3,2) = 3

Click here to see Combination Math for C(3,2)__Simplifying the rest of our expansion, we get:__4C(3,2)x

^{1}y

^{2} = (4)C(3,2)(2a

^{7})

^{3 - 2}(-6b

^{4})

^{2}4C(3,2)x

^{1}y

^{2} = (4)(3)(2a

^{7})

^{1}(-6b

^{4})

^{2}4C(3,2)x

^{1}y

^{2} = (4)(3)(2a

^{7})(36b

^{8})

__Group constants and powers from all terms__4C(3,2)x

^{1}y

^{2} = (4 * 3 * 2 * 36)(a

^{7}b

^{8})

4C(3,2)x

^{1}y

^{2} = 864a

^{7}b

^{8}

__k = 3 expansion:__4C(3,3)x

^{3-3}y

^{3} = 4C(3,3)x

^{0}y

^{3}__With n = 3 and k = 3, C(3,3) becomes:__C(3,3) = 1

Click here to see Combination Math for C(3,3)__Simplifying the rest of our expansion, we get:__4C(3,3)x

^{0}y

^{3} = (4)C(3,3)(2a

^{7})

^{3 - 3}(-6b

^{4})

^{3}4C(3,3)x

^{0}y

^{3} = (4)(1)(2a

^{7})

^{0}(-6b

^{4})

^{3}4C(3,3)x

^{0}y

^{3} = (4)(1)(1)(-216b

^{12})

__Group constants and powers from all terms__4C(3,3)x

^{0}y

^{3} = (4 * 1 * 1 * -216)(b

^{12})

4C(3,3)x

^{0}y

^{3} = -864b

^{12}

__Our resulting expanded term using the binomial theorem becomes:__4(2a

^{7} - 6b

^{4})

^{3} = 32a

^{21} - 288a

^{14}b

^{4} + 864a

^{7}b

^{8} - 864b

^{12}************ End Binomial Expansion ********************