we have \sin\, 54^\circ = \cos \, 36^\circ
let x = 18^\circ
so we have \sin 3x = \cos 2x
Or 3 \sin \, x - 4\sin ^3 x = 1 - 2 \sin ^2 x
putting \sin \,x = y we get
3y - 4y^3 = 1 - 2y^2
or 4y^3 - 2y^2 - 3 y + 1 = 0
or we see that (y-1) is a factor as putting above as f(y) and find f(1) = 0
by division we get (y-1)(4y^2 + 2y -1) = 0
as \sin\,18^\circ is not zero so 4y^2 + 2y - 1=0\cdots(1)
we need to show y^3 + y^2 = \frac{1}{8}
from (1) 4y^2 = - 2y + 1\cdots(2)
or 8y^3 = 2y(-2y + 1) = - 4y^2 + 2y = - -4y^2 + 1 - 4y^2 (from (2))
or 8y^3 + 8y^2 = 1
or y^3 + y^2 = \frac{1}{8}
proved