Cho (x+y)^2 - 2(x^2+y^2) = 0

Let's analyze each equation step-by-step.

### a) \((x+y)^2 - 2(x^2+y^2) = 0\)

We can rewrite the equation:

\[
(x+y)^2 = 2(x^2+y^2)
\]

Expanding the left-hand side:

\[
x^2 + 2xy + y^2 = 2(x^2 + y^2)
\]

This simplifies to:

\[
x^2 + 2xy + y^2 = 2x^2 + 2y^2
\]

Rearranging gives:

\[
2xy = 2x^2 + 2y^2 - x^2 - y^2
\]
\[
2xy = x^2 + y^2
\]

This can be rearranged to:

\[
x^2 - 2xy + y^2 = 0
\]
\[
(x - y)^2 = 0
\]

Thus, this implies:

\[
x = y
\]

### b) \((a+b+c)^2 = 3(ab+ac+bc)\)

We can expand the left-hand side:

\[
a^2 + b^2 + c^2 + 2(ab + ac + bc) = 3(ab + ac + bc)
\]

Rearranging gives:

\[
a^2 + b^2 + c^2 - ab - ac - bc = 0
\]

This can be rewritten as:

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

Since all squared terms are non-negative, this indicates that:

\[
a - b = 0 \quad \text{and} \quad b - c = 0 \quad \text{and} \quad c - a = 0
\]

Therefore, we conclude:

\[
a = b = c
\]

### c) \(a+b+c=0\)

Using the identity for sums of cubes, we have:

\[
a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc)
\]

Since we know \(a + b + c = 0\), we substitute that into the equation, which results in:

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

Thus, it simplifies to:

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

Now, we need to verify the condition \(3abc - (a^3 + b^3 + c^3) = 0\) holds, which is fulfilled by our earlier statement.

### Conclusion:
1. \(x = y\) from part a.
2. \(a = b = c\) from part b.
3. \(a^3 + b^3 + c^3 = 3abc\) holds for \(a + b + c = 0\) in part c.

Each condition establishes a relationship among the variables provided.