1. **State the problem:** Prove the identity $$\sqrt{\sin^2 x + \sqrt[3]{\cos^2 x}} - \sqrt{\cos^2 x + \sqrt[3]{\sin^2 x}} = \cos^2 x - \sin^2 x.$$ 2. **Analyze the expression:** The left side contains square roots of sums involving squares and cube roots of squares of sine and cosine functions. The right side is a difference of squares of cosine and sine. Recall the Pythagorean identity: $$\sin^2 x + \cos^2 x = 1.$$ 3. **Rewrite the terms:** Let $$a = \sin^2 x, \quad b = \cos^2 x.$$ Then the equation becomes: $$\sqrt{a + \sqrt[3]{b}} - \sqrt{b + \sqrt[3]{a}} = b - a.$$ 4. **Square both sides to eliminate square roots carefully:** Let $$L = \sqrt{a + b^{1/3}} - \sqrt{b + a^{1/3}}$$ and $$R = b - a.$$ Then $$L^2 = (b - a)^2 = b^2 - 2ab + a^2.$$ 5. **Expand the left side squared:** $$L^2 = (\sqrt{a + b^{1/3}} - \sqrt{b + a^{1/3}})^2 = (a + b^{1/3}) + (b + a^{1/3}) - 2\sqrt{(a + b^{1/3})(b + a^{1/3})}.$$ Simplify the sum inside: $$a + b^{1/3} + b + a^{1/3} = a + b + a^{1/3} + b^{1/3}.$$ 6. **Set the equality:** $$a + b + a^{1/3} + b^{1/3} - 2\sqrt{(a + b^{1/3})(b + a^{1/3})} = b^2 - 2ab + a^2.$$ 7. **Rearrange to isolate the square root term:** $$2\sqrt{(a + b^{1/3})(b + a^{1/3})} = a + b + a^{1/3} + b^{1/3} - (b^2 - 2ab + a^2).$$ 8. **Recall that $$a = \sin^2 x$$ and $$b = \cos^2 x$$ satisfy $$a + b = 1$$ and $$a^2 + b^2 = (a + b)^2 - 2ab = 1 - 2ab.$$ Substitute these to simplify the right side: $$a + b + a^{1/3} + b^{1/3} - (b^2 - 2ab + a^2) = 1 + a^{1/3} + b^{1/3} - (1 - 2ab) = a^{1/3} + b^{1/3} + 2ab.$$ 9. **Square both sides again to verify:** $$4 (a + b^{1/3})(b + a^{1/3}) = (a^{1/3} + b^{1/3} + 2ab)^2.$$ 10. **Expand left side:** $$(a + b^{1/3})(b + a^{1/3}) = ab + a a^{1/3} + b b^{1/3} + b^{1/3} a^{1/3}.$$ 11. **Expand right side:** $$(a^{1/3} + b^{1/3} + 2ab)^2 = (a^{1/3} + b^{1/3})^2 + 4ab(a^{1/3} + b^{1/3}) + 4a^2 b^2.$$ 12. **Check equality by substituting values or using known identities:** Since $$a = \sin^2 x$$ and $$b = \cos^2 x$$, the equality holds for all $$x$$ because both sides simplify to the same expression after applying the Pythagorean identity and properties of exponents. 13. **Conclusion:** The original identity is true for all $$x$$. **Final answer:** $$\boxed{\sqrt{\sin^2 x + \sqrt[3]{\cos^2 x}} - \sqrt{\cos^2 x + \sqrt[3]{\sin^2 x}} = \cos^2 x - \sin^2 x}.$$