When your calculator doesn't help

Let $\displaystyle x=3.25^\circ$ , then the expression becomes:

$\displaystyle\sin^{6}x + \cos^{6}x + 3\sin^{2}x\cos^{2}x$

= $[(\displaystyle\sin^{2}x)^{3}+(\cos^{2}x)^{3}]+3\sin^{2}x\cos^{2}x$

= $\displaystyle(\sin^{2}x+\cos^{2}x)(\sin^{4}x-\sin^{2}x\cos^{2}x+\cos^{4}x)+3\sin^{2}x\cos^{2}x$

= $\displaystyle\sin^{4}x-\sin^{2}x\cos^{2}x+\cos^{4}x+3\sin^{2}x\cos^{2}x$

= $\displaystyle\sin^{4}x+2\sin^{2}x\cos^{2}x+\cos^{4}x$

= $\displaystyle(\sin^{2}x)^{2}+(\cos^{2}x)^{2}+2.\sin^{2}x.\cos^{2}x$

= $\displaystyle(\sin^{2}x+\cos^{2}x)^{2}=1^{2}=\boxed{1}$

Hence, we find that $\displaystyle\sin^{6}x + \cos^{6}x + 3\sin^{2}x\cos^{2}x=1$ for every legitimate value of $x$ . So, $\displaystyle\sin^{6}x + \cos^{6}x + 3\sin^{2}x\cos^{2}x=1$ is rather an identity.

We can solve it by realising $a^3+b^3=(a+b)(a^2-ab+b^2)$ where $a=\sin^2(x)$ and $b=\cos^2(x)$

Consider the following:

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

Therefore, $\sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x = \boxed{1}$ for all $x$ including $3.25^\circ$ .