Let's solve the given expressions step-by-step.

Part 1: Show that $(a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a) = 0$

We can expand each term:

1. Expand each product: $(a - b)(a + b) = a^2 - b^2$ $(b - c)(b + c) = b^2 - c^2$ $(c - a)(c + a) = c^2 - a^2$

2. Combine the results: $(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2)$

3. Group the terms: $a^2 - b^2 + b^2 - c^2 + c^2 - a^2$

4. Observe that each term cancels: $(a^2 - a^2) + (b^2 - b^2) + (c^2 - c^2) = 0$

Thus, we have shown that: $(a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a) = 0$

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Part 2: Subtract $3a(a + b + c) - 2b(a - b + c)$ from $4c(-a + b + c)$

1. Expand each term:

   • For $4c(-a + b + c)$: $4c(-a) + 4c(b) + 4c(c) = -4ac + 4bc + 4c^2$
   • For $3a(a + b + c)$: $3a^2 + 3ab + 3ac$
   • For $-2b(a - b + c)$: $-2ba + 2b^2 - 2bc$

2. Combine the expression for the subtraction: $4c(-a + b + c) - (3a(a + b + c) - 2b(a - b + c))$ This translates to: $(-4ac + 4bc + 4c^2) - (3a^2 + 3ab + 3ac - 2ba + 2b^2 - 2bc)$

3. Distributing the negative sign: $-4ac + 4bc + 4c^2 - 3a^2 - 3ab - 3ac + 2ba - 2b^2 + 2bc$

4. Combine like terms:

   • The $a^2$ terms: $-3a^2$
   • The $b^2$ terms: $4bc + 2bc - 2b^2 = 6bc - 2b^2$
   • The $c^2$ terms: $4c^2$
   • The $ab$ terms: $-3ab + 2ba = -ab$
   • The $ac$ terms: $-4ac - 3ac = -7ac$

So, the final result is: $-3a^2 + (6bc - 2b^2) + 4c^2 - ab - 7ac$

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Would you like more details on any part of this solution? Here are some follow-up questions:

1. How can you verify the identity using specific numerical values for $a$, $b$, and $c$?
2. What are some properties of polynomial identities that help simplify these expressions?
3. Can you derive similar identities for three variables?
4. How do these algebraic manipulations relate to geometric interpretations?
5. What methods can be used to check the correctness of algebraic identities?

Tip: When dealing with polynomial identities, systematically expand and simplify each term to look for cancellations or patterns.