You need to use the binomial formula, such that:
`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k`
You need to replace y for x, 4 for y and 3 for n, such that:
`(y-4)^3 = 3C0 (y)^3 + 3C1 (y)^2*(-4)^1 + 3C2 (y)^1*(-4)^2 + 3C3(-4)^3 `
By definition, nC0 = nCn = 1, hence `3C0 = 3C3 = 1.`
By definition nC1 = nC(n-1) = n, hence `3C1 = 3C2 = 3.`
`(y-4)^3 = (y)^3 + 3(y)^2*(-4)^1 +3 (y)^1*16- 64 `
`(y-4)^3 = (y)^3- 12(y)^2 +48(y)^1- 64 `
Hence, expanding the complex number using binomial theorem yields the simplified result `(y-4)^3 = (y)^3- 12(y)^2 +48(y)^1- 64.`
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