Large Powered Trigonometric Functions

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

Relevant wiki: Pythagorean Identities

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

$\sin^6\theta + \cos^6\theta + 3\sin^2\theta\cos^2\theta = x \\ \implies \sin^6\theta + \cos^6\theta + 3\sin^2\theta\cos^2\theta\color{#3D99F6}{\times1} = x \\ \sin^6\theta + \cos^6\theta + 3\sin^2\theta\cos^2\theta\color{#3D99F6}{(\cos^2\theta+\sin^2\theta)} = x \\ \text{Let }\sin^2\theta \text{ be 'a' and }\cos^2\theta\text{ be 'b', Then} \\ x = a^3 + b^3 + 3ab(a+b) \\ x = (a+b)^3 \\ x = (\color{#D61F06}{cos^2\theta + sin^2\theta})^2 \\ x = \color{#D61F06}{1}^2 \\ \fbox{ x = 1 }$

Let sin^2(theta)=m ,so cos^2(theta)=1-m. Substitute in our expression ,m^3+(1-m)^3+3m(1-m)=m^3+1+m^2+m-m-m^3-m^2=1

nice solution.

Simply put theta =0 if you want to answer the question only.