Expanding the left side of the equation:

(y-3)^3 = (y-3)(y-3)(y-3)
= (y^2 - 6y + 9)(y-3)
= y^3 - 6y^2 + 9y - 3y^2 + 18y - 27
= y^3 - 9y^2 + 27y - 27

Expanding the term 2y(5y + 1):

2y(5y + 1) = 10y^2 + 2y

Adding the two expanded terms together:

(y-3)^3 + 2y(5y+1) = y^3 - 9y^2 + 27y - 27 + 10y^2 + 2y
= y^3 + 10y^2 - 9y^2 + 27y + 2y - 27
= y^3 + y^2 + 29y - 27

Expanding the right side of the equation:

y^3 - (2y - 1)^2 - 26
= y^3 - (4y^2 - 4y + 1) - 26
= y^3 - 4y^2 + 4y - 1 - 26
= y^3 - 4y^2 + 4y - 27

Therefore, the equation becomes:
y^3 + y^2 + 29y - 27 = y^3 - 4y^2 + 4y - 27

Simplifying further:
y^2 + 29y = - 4y^2 + 4y
5y^2 - 25y = 0
5y(y - 5) = 0

So, the solution is y = 0 and y = 5.