2022/032) Show that $\sin ^3 18^\circ + \sin ^2 18^\circ = \frac{1}{8}$

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