You need first to factorize `(4x-1)^3` , such that:
`(4x-1)^3 - 2(4x-1)^4 = (4x-1)^3(1 - 2(4x-1))`
`(4x-1)^3 - 2(4x-1)^4 = (4x-1)^3(1 - 8x + 2) `
`(4x-1)^3 - 2(4x-1)^4 = (4x-1)^3(3-8x)`
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 4x for x, 1 for y and 3 for n, such that:
`(4x-1)^3 = 3C0 (4x)^3 +3C1 (4x)^2*(-1)^1+3C2 (4x)*(-1)^2 + 3C3 (-1)^3`
By definition, nC0 = nCn = 1, hence `3C0 = 3C3 =...
You need first to factorize `(4x-1)^3` , such that:
`(4x-1)^3 - 2(4x-1)^4 = (4x-1)^3(1 - 2(4x-1))`
`(4x-1)^3 - 2(4x-1)^4 = (4x-1)^3(1 - 8x + 2) `
`(4x-1)^3 - 2(4x-1)^4 = (4x-1)^3(3-8x)`
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 4x for x, 1 for y and 3 for n, such that:
`(4x-1)^3 = 3C0 (4x)^3 +3C1 (4x)^2*(-1)^1+3C2 (4x)*(-1)^2 + 3C3 (-1)^3`
By definition, nC0 = nCn = 1, hence `3C0 = 3C3 = 1.`
By definition nC1 = nC(n-1) = n, hence `3C1= 3C2 = 3.`
`(4x-1)^3 = 64x^3 - 48x^2 + 12x - 1`
Replacing the binomial expansion `64x^3 - 48x^2 + 12x - 1` for `(4x-1)^3` yields:
`(4x-1)^3(3-8x) = (64x^3 - 48x^2 + 12x - 1)(3-8x)`
`(4x-1)^3(3-8x) = (192x^3 - 512x^4 - 144x^2 + 384x^3 + 36x - 96x^2 - 3 + 8x)`
Grouping the like terms yields:
`(4x-1)^3(3-8x) = (- 512x^4 + 576x^3 - 240x^2 + 44x - 3)`
Hence, expanding and simplifying the expression yields `(4x-1)^3 - 2(4x-1)^4 = (- 512x^4 + 576x^3 - 240x^2 + 44x - 3).`
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