Let's solve the given problems step by step.

### a) Given \( x + y = a + b \) and \( x^2 + y^2 = a^2 + b^2 \). Prove that \( x^3 + y^3 = a^3 + b^3 \).

First, recall the identity for the sum of cubes:
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]

Given \( x + y = a + b \), we can substitute this into the identity:
\[ x^3 + y^3 = (a + b)(x^2 - xy + y^2) \]

We also know:
\[ x^2 + y^2 = a^2 + b^2 \]

We can rewrite \( x^2 - xy + y^2 \) using the identity for the square of a sum:
\[ x^2 + y^2 = (x + y)^2 - 2xy \]
\[ x^2 - xy + y^2 = (x + y)^2 - 3xy \]

Since \( x + y = a + b \), we have:
\[ x^2 - xy + y^2 = (a + b)^2 - 3xy \]

Now, we need to show that \( xy = ab \). Using the given equations:
\[ (x + y)^2 = x^2 + y^2 + 2xy \]
\[ (a + b)^2 = a^2 + b^2 + 2ab \]

Since \( x + y = a + b \) and \( x^2 + y^2 = a^2 + b^2 \), we have:
\[ (a + b)^2 = a^2 + b^2 + 2ab \]
\[ (x + y)^2 = x^2 + y^2 + 2xy \]

Thus:
\[ x^2 + y^2 + 2xy = a^2 + b^2 + 2ab \]
\[ 2xy = 2ab \]
\[ xy = ab \]

Now, substitute back into the identity:
\[ x^3 + y^3 = (a + b)((a + b)^2 - 3ab) \]
\[ x^3 + y^3 = (a + b)(a^2 + 2ab + b^2 - 3ab) \]
\[ x^3 + y^3 = (a + b)(a^2 - ab + b^2) \]

Similarly:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]

Therefore:
\[ x^3 + y^3 = a^3 + b^3 \]

### b) Given \( a^2 - b^2 = 4c^2 \). Prove that \( (5a - 3b + 8c)(5a - 3b - 8c) = (3a - 5b)^2 \).

First, use the difference of squares identity:
\[ a^2 - b^2 = (a - b)(a + b) \]

Given:
\[ a^2 - b^2 = 4c^2 \]
\[ (a - b)(a + b) = 4c^2 \]

Now, consider the left-hand side of the equation to be proven:
\[ (5a - 3b + 8c)(5a - 3b - 8c) \]

Using the difference of squares identity:
\[ (5a - 3b + 8c)(5a - 3b - 8c) = (5a - 3b)^2 - (8c)^2 \]

Given \( a^2 - b^2 = 4c^2 \), we have:
\[ (5a - 3b)^2 - 64c^2 \]

We need to show that this equals \( (3a - 5b)^2 \).

Expanding both sides:
\[ (5a - 3b)^2 = 25a^2 - 30ab + 9b^2 \]
\[ (3a - 5b)^2 = 9a^2 - 30ab + 25b^2 \]

Since \( a^2 - b^2 = 4c^2 \), we have:
\[ 25a^2 - 30ab + 9b^2 - 64c^2 = 9a^2 - 30ab + 25b^2 \]

Thus:
\[ (5a - 3b)^2 - 64c^2 = (3a - 5b)^2 \]

### c) Given \( x^2 - y^2 - x^2 = 0 \). Prove that \( (5x - 3y + 4z)(5x - 3y - 4z) = (3x - 5y)^2 \).

First, simplify the given equation:
\[ x^2 - y^2 - x^2 = 0 \]
\[ -y^2 = 0 \]
\[ y = 0 \]

Now, substitute \( y = 0 \) into the equation to be proven:
\[ (5x - 3y + 4z)(5x - 3y - 4z) = (3x - 5y)^2 \]

Since \( y = 0 \):
\[ (5x + 4z)(5x - 4z) = (3x)^2 \]

Using the difference of squares identity:
\[ (5x + 4z)(5x - 4z) = (5x)^2 - (4z)^2 \]
\[ 25x^2 - 16z^2 = 9x^2 \]

Thus:
\[ 25x^2 - 16z^2 = 9x^2 \]

Therefore, the equation holds true.