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(2a²b³c⁴ - 6x³y⁴z⁵)⁵

Binomial Expansion

Since (2a²b³c⁴ - 6x³y⁴z⁵)⁵ is a binomial
Use the binomial theorem to expand this.

Build binomial expansion

a(x + y)ⁿ = aΣ_{[k = 0 to n]}C(n,k) x^n-ky^k
where

Plug in values

n = 5, x = 2a²b³c⁴, a = 1, and y = -6x³y⁴z⁵.
Expanding terms, we get:

k = 0

C(5,0)x^5-0y⁰ = C(5,0)x⁵y⁰

C(5,0):

C(5,0) = 1
Click to see C(5,0)

Simplify our expanded term:

C(5,0)x⁵y⁰ = C(5,0)(2a²b³c⁴)^{5 - 0}(-6x³y⁴z⁵)⁰

C(5,0)x⁵y⁰ = (1)(2a²b³c⁴)⁵(-6x³y⁴z⁵)⁰

C(5,0)x⁵y⁰ = (1)(1)(32a¹⁰b¹⁵c²⁰)(1)

Group constants and powers

C(5,0)x⁵y⁰ = (1 * 1 * 32 * 1)(1)
Anything raised to a 0 power = 1

C(5,0)x⁵y⁰ = 32a¹⁰b¹⁵c²⁰

k = 1

C(5,1)x^5-1y¹ = C(5,1)x⁴y¹

C(5,1):

C(5,1) = 5
Click to see C(5,1)

Simplify our expanded term:

C(5,1)x⁴y¹ = C(5,1)(2a²b³c⁴)^{5 - 1}(-6x³y⁴z⁵)¹

C(5,1)x⁴y¹ = (5)(2a²b³c⁴)⁴(-6x³y⁴z⁵)¹

C(5,1)x⁴y¹ = (1)(5)(16a⁸b¹²c¹⁶)(-6x³y⁴z⁵)

Group constants and powers

C(5,1)x⁴y¹ = (1 * 5 * 16 * -6)(a⁸b¹²c¹⁶x³y⁴z⁵)

C(5,1)x⁴y¹ = -480a⁸b¹²c¹⁶x³y⁴z⁵

k = 2

C(5,2)x^5-2y² = C(5,2)x³y²

C(5,2):

C(5,2) = 10
Click to see C(5,2)

Simplify our expanded term:

C(5,2)x³y² = C(5,2)(2a²b³c⁴)^{5 - 2}(-6x³y⁴z⁵)²

C(5,2)x³y² = (10)(2a²b³c⁴)³(-6x³y⁴z⁵)²

C(5,2)x³y² = (1)(10)(8a⁶b⁹c¹²)(36x⁶y⁸z¹⁰)

Group constants and powers

C(5,2)x³y² = (1 * 10 * 8 * 36)(a⁶b⁹c¹²x⁶y⁸z¹⁰)

C(5,2)x³y² = 2880a⁶b⁹c¹²x⁶y⁸z¹⁰

k = 3

C(5,3)x^5-3y³ = C(5,3)x²y³

C(5,3):

C(5,3) = 10
Click to see C(5,3)

Simplify our expanded term:

C(5,3)x²y³ = C(5,3)(2a²b³c⁴)^{5 - 3}(-6x³y⁴z⁵)³

C(5,3)x²y³ = (10)(2a²b³c⁴)²(-6x³y⁴z⁵)³

C(5,3)x²y³ = (1)(10)(4a⁴b⁶c⁸)(-216x⁹y¹²z¹⁵)

Group constants and powers

C(5,3)x²y³ = (1 * 10 * 4 * -216)(a⁴b⁶c⁸x⁹y¹²z¹⁵)

C(5,3)x²y³ = -8640a⁴b⁶c⁸x⁹y¹²z¹⁵

k = 4

C(5,4)x^5-4y⁴ = C(5,4)x¹y⁴

C(5,4):

C(5,4) = 5
Click to see C(5,4)

Simplify our expanded term:

C(5,4)x¹y⁴ = C(5,4)(2a²b³c⁴)^{5 - 4}(-6x³y⁴z⁵)⁴

C(5,4)x¹y⁴ = (5)(2a²b³c⁴)¹(-6x³y⁴z⁵)⁴

C(5,4)x¹y⁴ = (1)(5)(2a²b³c⁴)(1296x¹²y¹⁶z²⁰)

Group constants and powers

C(5,4)x¹y⁴ = (1 * 5 * 2 * 1296)(a²b³c⁴x¹²y¹⁶z²⁰)

C(5,4)x¹y⁴ = 12960a²b³c⁴x¹²y¹⁶z²⁰

k = 5

C(5,5)x^5-5y⁵ = C(5,5)x⁰y⁵

C(5,5):

C(5,5) = 1
Click to see C(5,5)

Simplify our expanded term:

C(5,5)x⁰y⁵ = C(5,5)(2a²b³c⁴)^{5 - 5}(-6x³y⁴z⁵)⁵

C(5,5)x⁰y⁵ = (1)(2a²b³c⁴)⁰(-6x³y⁴z⁵)⁵

C(5,5)x⁰y⁵ = (1)(1)(1)(-7776x¹⁵y²⁰z²⁵)

Group constants and powers

C(5,5)x⁰y⁵ = (1 * 1 * 1 * -7776)(x¹⁵y²⁰z²⁵)

C(5,5)x⁰y⁵ = -7776x¹⁵y²⁰z²⁵

Build our answer:

(2a²b³c⁴ - 6x³y⁴z⁵)⁵ = 32a¹⁰b¹⁵c²⁰ - 480a⁸b¹²c¹⁶x³y⁴z⁵ + 2880a⁶b⁹c¹²x⁶y⁸z¹⁰ - 8640a⁴b⁶c⁸x⁹y¹²z¹⁵ + 12960a²b³c⁴x¹²y¹⁶z²⁰ - 7776x¹⁵y²⁰z²⁵

************ End Binomial Expansion ********************

Final Answer