If `a^(3) + b^(3) + c^(3)=3abc` and a, b , c are distinct numbers. Which option is correct?
(a) `a+b+c= 0` (b) a= b= c

Answer

To solve the problem, we start with the equation given:

**Step 1: Write down the equation.**
We have:
\[ a^3 + b^3 + c^3 = 3abc \]

**Step 2: Recall the identity for cubes.**
There is a well-known algebraic identity that states:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

**Step 3: Apply the identity to our equation.**
From the identity, we can rearrange it as follows:
\[ a^3 + b^3 + c^3 - 3abc = 0 \]
This implies:
\[ (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) = 0 \]

**Step 4: Analyze the factors.**
For the product to be zero, at least one of the factors must be zero. Thus, we have two cases:
1. \( a + b + c = 0 \)
2. \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \)

**Step 5: Consider the distinct numbers condition.**
Given that \( a, b, c \) are distinct numbers, we analyze the second case:
- The equation \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \) simplifies to:
\[ (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 \]
This implies that \( a = b = c \), which contradicts the condition that \( a, b, c \) are distinct.

**Step 6: Conclude the solution.**
Since the second case cannot hold true, we must have:
\[ a + b + c = 0 \]

Thus, the correct option is:
**(a) \( a + b + c = 0 \)**

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