Precalculus, Chapter 9, 9.5, Section 9.5, Problem 40

We have to expand the expression using the binomial theorem , so use the formula,
(a+b)^n=sum_(k=0)^n((n),(k))a^(n-k)b^k
(4x-1)^3=((3),(0))(4x)^(3-0)(-1)^0+((3),(1))(4x)^(3-1)(-1)^1+((3),(2))(4x)^(3-2)(-1)^2+((3),(3))(4x)^(3-3)(-1)^3
=(4x)^3+(3!)/(1!(3-1)!)(4x)^2(-1)+(3!)/(2!(3-2)!)(4x)^1(1)+(-1)
=64x^3+(3*2!)/(2!)(-16)x^2+(3*2!)/(2!)(4x)-1
=64x^3-48x^2+12x-1
(4x-1)^4=((4),(0))(4x)^(4-0)(-1)^0+((4),(1))(4x)^(4-1)(-1)^1+((4),(2))(4x)^(4-2)(-1)^2+((4),(3))(4x)^(4-3)(-1)^3+((4),(4))(4x)^(4-4)(-1)^4
=(4x)^4+(4!)/(1!(4-1)!)(4x)^3(-1)+(4!)/(2!(4-2)!)(4x)^2(1)+(4!)/(3!(4-3)!)(4x)^1(-1)+(-1)^4
=256x^4+(4*3!)/(1!3!)64x^3(-1)+(4*3*2!)/(2!2!)16x^2+(4*3!)/(3!)(4x)(-1)+1
=256x^4-256x^3+96x^2-16x+1
:.(4x-1)^3-2(4x-1)^4=(64x^3-48x^2+12x-1)-2(256x^4-256x^3+96x^2-16x+1)
=64x^3-48x^2+12x-1-512x^4+512x^3-192x^2+32x-2
=-512x^4+64x^3+512x^3-48x^2-192x^2+12x+32x-1-2
=-512x^4+576x^3-240x^2+44x-3