An algebra problem by Suresh Palaparthi

$\sin^{6} \theta + \cos^{6} \theta +3 \sin^{2} \theta \times \cos^{2} \theta$ $= \sin^{6} \theta + \cos^{6} \theta +3 \sin^{2} \theta \times \cos^{2} \theta \times 1$ $= \sin^{6} \theta + \cos^{6} \theta +3 \sin^{2} \theta \times \cos^{2} \theta \times (sin^2 \theta + cos^2 \theta)$

Recall identity : $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$

Therefore

$= \sin^{6} \theta + \cos^{6} \theta +3 \sin^{2} \theta \times \cos^{2} \theta \times (sin^2 \theta + cos^2 \theta)$ $= (sin^2 \theta + cos^2 \theta)^3$ $= \boxed{1}$

$\begin{aligned} X & = \sin^6 \theta +\cos^2 \theta + 3 \sin^2 \theta \cos^2 \theta \\ & = {\color{#3D99F6}\left(\sin^2 \theta + \cos^2 \theta \right)}^3 - 3\sin^2 \theta \cos^2 \theta {\color{#3D99F6}\left(\sin^2 \theta + \cos^2 \theta \right)} + 3 \sin^2 \theta \cos^2 \theta \\ & = {\color{#3D99F6}1}^3 - 3\sin^2 \theta \cos^2 \theta {\color{#3D99F6}\left(1\right)} + 3 \sin^2 \theta \cos^2 \theta \\ & = \boxed{1} \end{aligned}$