Calculations with Trig

$\begin{aligned} x & = 2{\color{#3D99F6}(\sin^6 \theta + \cos^6 \theta)} - 3 (\sin^4 \theta + \cos^4 \theta) & \small \color{#3D99F6} \text{Note that }a^3+b^3 = (a+b)(a^2+b^2-ab) \\ & = 2{\color{#3D99F6}({\color{#D61F06}\sin^2 \theta + \cos^2 \theta}) (\sin^4 \theta + \cos^4 \theta - \sin^2 \theta \cos^2 \theta)} - 3(\sin^4 \theta + \cos^4 \theta) & \small \color{#D61F06} \text{Note that }\sin^2 x+\cos^2 x = 1 \\ & = - {\color{#3D99F6}(\sin^4 \theta + \cos^4 \theta)} - 2\sin^2 \theta \cos^2 \theta & \small \color{#3D99F6} \text{Note that }a^2+b^2 = (a+b)^2 -2ab \\ & = - \left(({\color{#D61F06}\sin^2 \theta + \cos^2 \theta})^2 - 2\sin^2 \theta \cos^2 \theta \right) - 2\sin^2 \theta \cos^2 \theta \\ & = \boxed{-1} \end{aligned}$

Hee hee. Since it can be assumed the value is constant for all $\theta$ , we can substitute $\theta=0$ , noting $\sin^n(0) = 0, \forall n$ and $\cos^n(0) = 1, \forall n$ . So $2(\sin^6\theta + \cos^6\theta)-3(\sin^4\theta+\cos^4\theta) \\ =2(0+1)-3(0+1) \\ =2-3 \\ =\boxed{-1}$

$\begin{aligned}2(\sin^6 \theta + \cos^6 \theta ) - 3(\sin^4 \theta +\cos^4 \theta) &= 2(\sin^2 \theta + \cos^2 \theta )(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) - 3(\sin^4 \theta +\cos^4 \theta) \\ &= 2(1)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta) - 3(\sin^4 \theta +\cos^4 \theta)\\ &= 2(\sin^4 \theta +\cos^4 \theta) - 2 \sin^2 \theta \cos^2 \theta - 3(\sin^4 \theta +\cos^4 \theta) \\ &= -(\sin^4 \theta +\cos^4 \theta) - 2 \sin^2 \theta \cos^2 \theta \\ &= -\sin^4 \theta - 2 \sin^2 \theta \cos^2 \theta - \cos^4 \theta \\ &= -(\sin^2 \theta + \cos^2 \theta)^2 \\ &= -1^2 \\ &= \boxed{-1} \end{aligned}$