Prove that sin6θ + cos6θ = cos4θ + sin4θ

The RHS is $\cos^2 \theta + \sin^2 \theta$.
Using the fundamental Pythagorean identity, $\sin^2 \theta + \cos^2 \theta = 1$.

$$\text{RHS} = 1$$

The LHS is $\sin^6 \theta + \cos^6 \theta$.
We can rewrite this expression using the sum of cubes formula, $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$.
Let $a = \sin^2 \theta$ and $b = \cos^2 \theta$.
Then, $\sin^6 \theta + \cos^6 \theta = (\sin^2 \theta)^3 + (\cos^2 \theta)^3$.

$$\text{LHS} = (\sin^2 \theta + \cos^2 \theta)((\sin^2 \theta)^2 - (\sin^2 \theta)(\cos^2 \theta) + (\cos^2 \theta)^2)$$

$$\text{LHS} = (1)(\sin^4 \theta - \sin^2 \theta \cos^2 \theta + \cos^4 \theta)$$ $$\text{LHS} = \sin^4 \theta + \cos^4 \theta - \sin^2 \theta \cos^2 \theta$$

We know that $(\sin^2 \theta + \cos^2 \theta)^2 = 1^2$.
Expanding the left side:

$$\sin^4 \theta + 2\sin^2 \theta \cos^2 \theta + \cos^4 \theta = 1$$

Rearranging this, we get:

$$\sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta$$

$$\text{LHS} = (1 - 2\sin^2 \theta \cos^2 \theta) - \sin^2 \theta \cos^2 \theta$$ $$\text{LHS} = 1 - 3\sin^2 \theta \cos^2 \theta$$

We have $\text{LHS} = 1 - 3\sin^2 \theta \cos^2 \theta$ and $\text{RHS} = 1$.
For the given equation to be true, we would need:

$$1 - 3\sin^2 \theta \cos^2 \theta = 1$$

Subtracting 1 from both sides:

$$-3\sin^2 \theta \cos^2 \theta = 0$$ $$\sin^2 \theta \cos^2 \theta = 0$$

This condition is only met when $\sin \theta = 0$ or $\cos \theta = 0$. This is not true for all values of $\theta$ (for example, if $\theta = \frac{\pi}{4}$, then $\sin \theta = \frac{\sqrt{2}}{2}$ and $\cos \theta = \frac{\sqrt{2}}{2}$, so $\sin^2 \theta \cos^2 \theta = \frac{1}{4} \neq 0$).

Therefore, the given statement is not a trigonometric identity and cannot be proven as generally true.

$$\boxed{\text{The statement is not a general identity.}}$$