TRIONGOMETRY

By Quadratic formula,

$\begin{aligned} \sin x & = & \frac {-1 + \sqrt 5} {2} \\ \sin^2 x & = & \frac {3- \sqrt 5}{2} \\ \cos^2 x & = & 1 - \sin^2 x = \frac {\sqrt5 - 1}{2} \\ \cos^6 x & = & ( \cos^2 x)^3 = \sqrt5 - 2 \\ \cos^{12} x & = & ( \cos^6 x)^2 = 9 - 4 \sqrt5 \\ \cos^{10} x & = & ( \cos^{12} x) \div \cos^2 x = \frac {5\sqrt5 - 11}{2} \\ \cos^8 x & = & ( \cos^{10} x) \div \cos^2 x = \frac {-3\sqrt5 +7}{2} \\ \end{aligned}$

Add up all the requird terms should give you $\boxed{0}$ as the answer.

given that sinx+(sinx)^2=1 then sinx=1-(sinx)^2=(cosx)^2 then substitute (cosx)^2 in the fining value. before that cubing on the given condition we get (sinx)^3+3(sinx)^4+3(sinx)^5+(sinx)^6=1 ((cosx)^2)^6+3((cosx)^2)^5+3((cosx)^2)^4+((cosx)^2)^3-1=(sinx)^3+3(sinx)^4+3(sinx)^5+(sinx)^6-1=(sinx+(sinx)^2)^3-(1) =1^3-1=0.