Easy Trigo.

Given $\sin\theta=1-\sin^2\theta=\cos^2\theta$ L.H.S $\cos^6\theta(\cos^2\theta+1)^3-1$ $=\sin^3\theta(\sin\theta+1)^3-1$ $=(\sin\theta(\sin\theta+1))^3-1$ $=(\sin^2\theta+\sin\theta)^3-1=1^3-1=\boxed0$

$Since\quad value\quad is\quad given\quad in\quad terms\quad of\quad sin\theta \quad I\quad thought\\ it\quad might\quad be\quad beneficial\quad to\quad convert\quad terms\quad of\quad cos\theta \quad in\quad sin\theta \\ moreover\quad its\quad apparent\quad that\quad it\quad can\quad be\quad done\quad easily.\quad \\ After\quad substituiting(\cos ^{ 2 }{ \theta } =\sin { \theta } )\quad I\quad got\quad these\quad terms:\\ \\ \sin ^{ 3 }{ \theta } +\sin ^{ 6 }{ \theta } +3\sin ^{ 4 }{ \theta } +3\sin ^{ 5 }{ \theta } -1\\ \\ Which\quad on\quad rearranging\quad gave\quad these:\\ \\ \Rightarrow \sin ^{ 3 }{ \theta } +\sin ^{ 6 }{ \theta } +3\sin ^{ 3 }{ \theta } (\sin ^{ 2 }{ \theta } +\sin { \theta } )-1\\ \\ I\quad finally\quad shot\quad the\quad problem\quad by\quad this:\\ \Rightarrow { (\sin { \theta } +\sin ^{ 2 }{ \theta } ) }^{ 3 }-1=0..$

I didn't factorize all the way...

I took sin( $\theta$ ) as X. X+ $X^2$ =1. After using the quadratic solver, X = [-1 $\pm$ $\sqrt5$ ]/2.

Also, sin( $\theta$ )+[sin( $\theta$ )]^2= cos( $\theta$ )+[sin( $\theta$ )]^2. sin( $\theta$ )+[cos( $\theta$ )]^2

The value becomes X^6+3X^5+3X^4+X^3-1=X^3(X+1)^3-1= {[-1 $\pm$ $\sqrt5$ ]/2}^3+{[1 $\pm$ $\sqrt5$ ]/2}^3-1

After some tiring calculations, I got 1-1=0. Argh... If I had realised...