We are given the equation:

$a^2 + b^2 + c^2 - ab - bc - ca = 0$

We need to prove that $a = b = c$.

Proof:

1. Rewrite the equation:

   The given equation can be rearranged as:

   $a^2 + b^2 + c^2 - ab - bc - ca = 0$

   This can also be expressed as:

   $\frac{1}{2} \left( 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca \right) = 0$

2. Use the identity:

   Recognize that the expression inside the parentheses can be factored using a well-known identity:

   $2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = (a - b)^2 + (b - c)^2 + (c - a)^2$

   Therefore, the equation becomes:

   $\frac{1}{2} \left( (a - b)^2 + (b - c)^2 + (c - a)^2 \right) = 0$

3. Interpret the equation:

   Since the sum of squares is zero, each individual square must be zero because the square of a real number is non-negative:

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

4. Solve the system:

   Solving each equation gives:

   $a - b = 0 \quad \Rightarrow \quad a = b$ $b - c = 0 \quad \Rightarrow \quad b = c$ $c - a = 0 \quad \Rightarrow \quad c = a$

   Therefore, we conclude:

   $a = b = c$

This completes the proof.

Would you like more details, or do you have any questions?

Here are some related questions to explore:

1. How does this identity simplify when dealing with complex numbers?
2. Can this approach be extended to vectors in three-dimensional space?
3. What happens if we allow non-real numbers for $a$, $b$, and $c$?
4. How does this identity relate to geometric representations of equilateral triangles?
5. Can this result be generalized for higher dimensions with similar equations?

Tip: Always look for symmetry in equations, as it often leads to simplifications in proofs.