I have two methods:
Using Calculus: Define $f(\theta)=\sin{\theta}\cos^5{\theta}-\sin^5{\theta}\cos{\theta}$. Now, we calculate:
$$\int{(f(\theta)\mathrm{d}\theta)}$$
in two ways. First, calculate it directly (using $u$-subs):
$$\int{(f(\theta)\mathrm{d}\theta)}=\int{((\sin{\theta}\cos^5{\theta}-\sin^5{\theta}\cos{\theta})\mathrm{d}\theta)}=-\frac{1}{6}(\cos^6{\theta}+\sin^6{\theta})+C_1$$
Second, simplify $f(\theta)$ first, and integrate afterwards:
$$f(\theta)=\sin{\theta}\cos^5{\theta}-\sin^5{\theta}\cos{\theta}=\sin{\theta}\cos{\theta}(\sin^4{\theta}-\cos^4{\theta})=\sin{\theta}\cos{\theta}(\sin^2{\theta}-\cos^2{\theta})(\sin^2{\theta}+\cos^2{\theta})=\sin{\theta}\cos{\theta}(\sin^2{\theta}-\cos^2{\theta})=\frac{1}{2}(2\sin{\theta}\cos{\theta})(-(\cos^2{\theta}-\sin^2{\theta}))=-\frac{1}{2}(\sin{2\theta})(\cos{2\theta})=-\frac{1}{4}(2\sin{2\theta}\cos{2\theta})=-\frac{1}{4}\sin{4\theta}$$
Now, we integrate $f(\theta)$:
$$\int{(f(\theta)\mathrm{d}\theta)}=\int{(-\frac{1}{4}\sin{4\theta}\mathrm{d}\theta)}=-\frac{1}{16}\cos{4\theta}+C_2$$
Of course, the point is that the two expressions are equal! i.e.:
$$-\frac{1}{6}(\cos^6{\theta}+\sin^6{\theta})+C_1=-\frac{1}{16}\cos{4\theta}+C_2$$
If we let $C_3$ be a single constant (instead of $C_1$ and $C_2$), and simplify:
$$\cos^6{\theta}+\sin^6{\theta}+C_3=\frac{3}{8}\cos{4\theta}$$
Plugging in a convenient value such as $\theta=0$ to find $C_3$:
$$\cos^6{0}+\sin^6{0}+C_3=\frac{3}{8}\cos{4(0)}$$
$$C_3=-\frac{5}{8}$$
Using the value of $C_3$:
$$\cos^6{\theta}+\sin^6{\theta}-\frac{5}{8}=\frac{3}{8}\cos{4\theta}$$
$$\cos^6{\theta}+\sin^6{\theta}=\frac{1}{8}(5+3\cos{4\theta})$$
Using Algebra: A slightly different way using algebra. We know that:
$$(x+y)^3=x^3+3x^2y+3xy^2+y^3=x^3+y^3+3xy(x+y)$$
so:
$$1=1^3=(\sin^2{\theta}+\cos^2{\theta})^3=\sin^6{\theta}+\cos^6{\theta}+3\sin^2{\theta}\cos^2{\theta}(\sin^2{\theta}+\cos^2{\theta})=\sin^6{\theta}+\cos^6{\theta}+3\sin^2{\theta}\cos^2{\theta}$$
$$1=\sin^6{\theta}+\cos^6{\theta}+3\sin^2{\theta}\cos^2{\theta}$$
$$\frac{3}{8}+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}+3\sin^2{\theta}\cos^2{\theta}$$
$$\frac{3}{8}-3\sin^2{\theta}\cos^2{\theta}+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$
$$\frac{3}{8}-\frac{3}{4}(4\sin^2{\theta}\cos^2{\theta})+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$
$$\frac{3}{8}-\frac{3}{4}(2\sin{\theta}\cos{\theta})^2+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$
$$\frac{3}{8}-\frac{3}{4}\sin^2{2\theta}+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$
$$\frac{3}{8}(1-2\sin^2{2\theta})+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$
$$\frac{3}{8}(\cos{4\theta})+\frac{5}{8}=\sin^6{\theta}+\cos^6{\theta}$$
$$\frac{1}{8}(5+3\cos{4\theta})=\sin^6{\theta}+\cos^6{\theta}$$
I hope you enjoy these two very elegant proofs!