Trigonometry for all

Substitute $\sin^{2}\theta = 1 - \cos^{2}\theta$ to get $\cos^{2}\theta -3\cos\theta + 1 =0 》》 \cos\theta(\cos \theta - 3 ) = -1 》》 \cos\theta -3 = -\frac {1}{\cos\theta} 》》 \cos\theta -3 = -\sec\theta 》》 \cos\theta + \sec\theta =3$ now cube both sides $\cos^{3}\theta + \sec^{3}\theta + 3 \cos\theta\sec\theta(\cos\theta + \sec\theta)= 27$ substitute $\cos\theta + \sec\theta = 3$ and $\sec\theta×\cos\theta =1$ to get the answer as $\boxed{18}$

Write a solution.

by substituting sin^2 as 1-cos^2, we get the quadratic equation, cos^2 - 3 cos + 1. solving it, we get cos as, {3+root(5)}/2 and cubing it twice for cos and sec, we get the value as 18.

My Solution: Using the $(a+b)^{ 3 }$ identity