1. The problem is to prove the identity $$a^3 + b^3 + c^3 = 3abc$$ under certain conditions. 2. This identity is true if and only if $$a + b + c = 0$$. This is a key condition to remember. 3. The formula for the sum of cubes with three variables is: $$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$ 4. If $$a + b + c = 0$$, then the right side becomes: $$(0)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$$ 5. Therefore, under the condition $$a + b + c = 0$$, we have: $$a^3 + b^3 + c^3 - 3abc = 0$$ which implies $$a^3 + b^3 + c^3 = 3abc$$ 6. This completes the proof. The key takeaway is that the identity holds when the sum of the three variables is zero. 7. To summarize: - Start with the known factorization. - Apply the condition $$a + b + c = 0$$. - Simplify to get the desired equality.