You need to consider `(2y – 1)^6 = (2y – 1)^3*(2y – 1)^3`

You need to expand the binomial `(2y – 1)^3` such that:

`(2y – 1)^3 = (2y)^3 - 3*(2y)^2*1 + 3*2y*1^2 - 1^3`

`(2y – 1)^3 = 8y^3 - 12y^2 + 6y - 1`

You need to multiply `8y^3 - 12y^2 + 6y - 1` by `8y^3 - 12y^2 + 6y - 1` such that:

`(2y – 1)^3*(2y – 1)^3 = (8y^3 - 12y^2 + 6y - 1)(8y^3 - 12y^2 + 6y - 1)`

`(2y – 1)^3*(2y – 1)^3 = 8y^3(8y^3 - 12y^2 + 6y - 1) - 12y^2 (8y^3 - 12y^2 + 6y - 1) + 6y(8y^3 - 12y^2 + 6y - 1) - (8y^3 - 12y^2 + 6y - 1` )

`(2y – 1)^3*(2y – 1)^3 = 64y^6 - 96y^5 + 48y^4 - 8y^3 - 96y^5 + 144y^4 - 72y^3 + 12y^2 + 48y^4 - 72y^3 + 36y^2 - 6y - 8y^3 + 12y^2 - 6y + 1`

Collecting like terms yields:

`(2y – 1)^3*(2y – 1)^3 = 64y^6- 192y^5 + 240y^4 - 160y^3 + 60y^2 - 12y + 1`

**Hence, evaluating `(2y – 1)^6` yields `(2y – 1)^6 = 64y^6- 192y^5 + 240y^4 - 160y^3 + 60y^2 - 12y + 1` .**