1. **State the problem:** Prove or verify the identity $$\sin^8 x - \cos^8 x = (\sin^2 x - \cos^2 x)(1 - 2\sin^2 x \cos^2 x)$$. 2. **Recall formulas and rules:** - Difference of powers: $$a^4 - b^4 = (a^2 - b^2)(a^2 + b^2)$$. - Pythagorean identity: $$\sin^2 x + \cos^2 x = 1$$. 3. **Rewrite left side:** Express $$\sin^8 x - \cos^8 x$$ as $$ (\sin^4 x)^2 - (\cos^4 x)^2 $$. 4. **Apply difference of squares:** $$ (\sin^4 x)^2 - (\cos^4 x)^2 = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x) $$. 5. **Factor $$\sin^4 x - \cos^4 x$$:** $$ \sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) $$. Using Pythagorean identity, $$\sin^2 x + \cos^2 x = 1$$, so $$ \sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x) \cdot 1 = \sin^2 x - \cos^2 x $$. 6. **Simplify $$\sin^4 x + \cos^4 x$$:** Rewrite as $$ (\sin^2 x)^2 + (\cos^2 x)^2 = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x $$. 7. **Combine all:** $$ \sin^8 x - \cos^8 x = (\sin^2 x - \cos^2 x)(1 - 2\sin^2 x \cos^2 x) $$. **Final answer:** The identity is verified and true.