To prove that \( \mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{c} = \mathbf{c} \times \mathbf{a} \) given that \( \mathbf{a} + \mathbf{b} + \mathbf{c} = 0 \), consider the properties of the cross product in vector algebra.

Given:
\[ \mathbf{a} + \mathbf{b} + \mathbf{c} = 0 \]

Rearrange the equation:
\[ \mathbf{a} + \mathbf{b} = -\mathbf{c} \]

Now let's take the cross product of both sides with \(\mathbf{a}\):
\[ \mathbf{a} \times (\mathbf{a} + \mathbf{b}) = \mathbf{a} \times (-\mathbf{c}) \]

Using the distributive property of the cross product:

\[ \mathbf{a} \times \mathbf{a} + \mathbf{a} \times \mathbf{b} = -\mathbf{a} \times \mathbf{c} \]

Since the cross product of any vector with itself is zero (\(\mathbf{a} \times \mathbf{a} = 0\)):

\[ 0 + \mathbf{a} \times \mathbf{b} = -\mathbf{a} \times \mathbf{c} \]

Therefore:

\[ \mathbf{a} \times \mathbf{b} = -\mathbf{a} \times \mathbf{c} \]

Next, recall another property of the cross product that negates this (- sign property):

\[ \mathbf{a} \times \mathbf{c} = -(\mathbf{c} \times \mathbf{a}) \]

Thus:

\[ \mathbf{a} \times \mathbf{b} = \mathbf{c} \times \mathbf{a} \]

Similarly, start with the initial condition and take the cross product with \(\mathbf{b}\):

\[ \mathbf{b} \times (\mathbf{a} + \mathbf{b}) = \mathbf{b} \times (-\mathbf{c}) \]

\[ \mathbf{b} \times \mathbf{a} + \mathbf{b} \times \mathbf{b} = -\mathbf{b} \times \mathbf{c} \]

Since \(\mathbf{b} \times \mathbf{b} = 0\):

\[ \mathbf{b} \times \mathbf{a} = -\mathbf{b} \times \mathbf{c} \]

Thus:

\[ -\mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{c} \]

Therefore:

\[ \mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{c} \]

Now, connect the results:

\[ \mathbf{a} \times \mathbf{b} = \mathbf{c} \times \mathbf{a} = \mathbf{b} \times \mathbf{c} \]

Thus, we've proven that:

\[ \mathbf{a} \times \mathbf{b} = \mathbf{b} \times \mathbf{c} = \mathbf{c} \times \mathbf{a} \]