1. **State the problem:** Verify the identity $$\sin^6 A + \cos^6 A + 3 \sin^2 A \cos^2 A = 1.$$\n\n2. **Recall important formulas:** We know that $$\sin^2 A + \cos^2 A = 1.$$ Also, the sum of cubes formula is useful: $$a^3 + b^3 = (a + b)^3 - 3ab(a + b).$$\n\n3. **Rewrite the expression:** Let $$x = \sin^2 A$$ and $$y = \cos^2 A.$$ Then the expression becomes $$x^3 + y^3 + 3xy.$$\n\n4. **Use the sum of cubes formula:** Since $$x + y = 1,$$ we have\n$$x^3 + y^3 = (x + y)^3 - 3xy(x + y) = 1^3 - 3xy \cdot 1 = 1 - 3xy.$$\n\n5. **Substitute back:** The original expression is\n$$x^3 + y^3 + 3xy = (1 - 3xy) + 3xy = 1.$$\n\n6. **Conclusion:** The identity holds true for all angles $$A$$ because it simplifies to 1.