Let a, b and c be real numbers such that a - 7b + 8c = 4 and 8a + 4b - c = 7. What is the value of `a^(2) - b^(2) + c^(2)`.

Answer

To solve the problem, we need to find the value of \( a^2 - b^2 + c^2 \) given the equations:

1. \( a - 7b + 8c = 4 \) (Equation 1)
2. \( 8a + 4b - c = 7 \) (Equation 2)

### Step 1: Rearranging the equations

We can rearrange both equations to express \( a \) and \( c \) in terms of \( b \).

From Equation 1:
\[ a + 8c = 4 + 7b \implies a = 4 + 7b - 8c \tag{1a} \]

From Equation 2:
\[ 8a - c = 7 - 4b \implies 8a = 7 - 4b + c \implies a = \frac{7 - 4b + c}{8} \tag{2a} \]

### Step 2: Substituting Equation 1a into Equation 2a

Now, substitute \( a \) from Equation 1a into Equation 2a:
\[ \frac{7 - 4b + c}{8} = 4 + 7b - 8c \]

### Step 3: Clear the fraction

Multiply both sides by 8 to eliminate the fraction:
\[ 7 - 4b + c = 32 + 56b - 64c \]

### Step 4: Rearranging the equation

Rearranging gives:
\[ c + 64c = 32 + 56b + 4b - 7 \]
\[ 65c = 25 + 60b \]
\[ c = \frac{25 + 60b}{65} = \frac{5 + 12b}{13} \tag{3} \]

### Step 5: Substitute \( c \) back into Equation 1

Now substitute \( c \) from Equation 3 back into Equation 1a to find \( a \):
\[ a = 4 + 7b - 8 \left(\frac{5 + 12b}{13}\right) \]
\[ = 4 + 7b - \frac{40 + 96b}{13} \]
\[ = \frac{52 + 91b - 40 - 96b}{13} \]
\[ = \frac{12 - 5b}{13} \tag{4} \]

### Step 6: Now we have \( a \) and \( c \) in terms of \( b \)

We have:
- \( a = \frac{12 - 5b}{13} \)
- \( c = \frac{5 + 12b}{13} \)

### Step 7: Finding \( a^2 - b^2 + c^2 \)

Now we can find \( a^2 - b^2 + c^2 \):
\[ a^2 = \left(\frac{12 - 5b}{13}\right)^2 = \frac{(12 - 5b)^2}{169} \]
\[ c^2 = \left(\frac{5 + 12b}{13}\right)^2 = \frac{(5 + 12b)^2}{169} \]

Now, substituting these into \( a^2 - b^2 + c^2 \):
\[ a^2 - b^2 + c^2 = \frac{(12 - 5b)^2 + (5 + 12b)^2 - 169b^2}{169} \]

### Step 8: Expanding and simplifying

Expanding:
\[ (12 - 5b)^2 = 144 - 120b + 25b^2 \]
\[ (5 + 12b)^2 = 25 + 120b + 144b^2 \]

Combining:
\[ = \frac{144 - 120b + 25b^2 + 25 + 120b + 144b^2 - 169b^2}{169} \]
\[ = \frac{169 + 0b + 0b^2}{169} = 1 \]

Thus, the value of \( a^2 - b^2 + c^2 \) is \( 1 \).

### Final Answer
\[ \boxed{1} \]