If a, b, c are real, `a^(3)+b^(3)+c^(3)=3abc and a+b+c ne 0`, then relation between a, b, c will be

Options

A. `a+b=c`
B. `a+c=b`
C. `a=b=c`
D. `b+c=a`

Answer

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

\[ a^3 + b^3 + c^3 = 3abc \]

and the condition that \(a + b + c \neq 0\).

### Step 1: Rewrite the equation
We can rearrange the equation as follows:

\[ a^3 + b^3 + c^3 - 3abc = 0 \]

### Step 2: Use the identity
There is a well-known algebraic identity that states:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c) \left( (a - b)^2 + (b - c)^2 + (c - a)^2 \right) \]

### Step 3: Analyze the equation
From the identity, we can see that the expression \(a^3 + b^3 + c^3 - 3abc = 0\) can be factored as:

\[ (a + b + c) \left( (a - b)^2 + (b - c)^2 + (c - a)^2 \right) = 0 \]

### Step 4: Consider the factors
Since we know that \(a + b + c \neq 0\), the first factor cannot be zero. Therefore, we must have:

\[ (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 \]

### Step 5: Solve the equation
The only way for the sum of squares to be zero is if each individual square is zero:

\[ (a - b)^2 = 0, \quad (b - c)^2 = 0, \quad (c - a)^2 = 0 \]

This implies:

\[ a - b = 0 \implies a = b \]
\[ b - c = 0 \implies b = c \]
\[ c - a = 0 \implies c = a \]

### Conclusion
Thus, we conclude that:

\[ a = b = c \]

### Final Result
The relation between \(a\), \(b\), and \(c\) is that they are all equal.

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