If x, y, z are in A.P., then `x^(3) + z^(3) - 8y^(3) = 4xyz`

Answer

To solve the problem, we need to show that if \( x, y, z \) are in arithmetic progression (A.P.), then the equation \( x^3 + z^3 - 8y^3 = 4xyz \) holds true.

### Step-by-Step Solution:

1. **Understanding A.P.**:
Since \( x, y, z \) are in A.P., we can express this relationship as:
\[ 2y = x + z \]
This means that \( y \) is the average of \( x \) and \( z \).

2. **Cubing the Terms**:
We want to manipulate the left-hand side (LHS) of the equation:
\[ LHS = x^3 + z^3 - 8y^3 \]

3. **Using the Identity for Cubes**:
We can use the identity for the sum of cubes:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Applying this to \( x^3 + z^3 \):
\[ x^3 + z^3 = (x + z)(x^2 - xz + z^2) \]
Since \( x + z = 2y \), we can substitute:
\[ x^3 + z^3 = 2y(x^2 - xz + z^2) \]

4. **Expressing \( y^3 \)**:
We also need to express \( 8y^3 \) in terms of \( x \) and \( z \):
\[ 8y^3 = 8\left(\frac{x + z}{2}\right)^3 = \frac{8}{8}(x + z)^3 = (x + z)^3 \]

5. **Expanding \( (x + z)^3 \)**:
Expanding \( (x + z)^3 \):
\[ (x + z)^3 = x^3 + z^3 + 3xz(x + z) \]
Therefore, we can rewrite \( 8y^3 \) as:
\[ 8y^3 = x^3 + z^3 + 3xz(2y) = x^3 + z^3 + 6xyz \]

6. **Substituting Back into LHS**:
Now substituting back into the LHS:
\[ LHS = x^3 + z^3 - 8y^3 = x^3 + z^3 - (x^3 + z^3 + 6xyz) = -6xyz \]

7. **Final Comparison**:
Now we compare the LHS with the RHS:
\[ LHS = -6xyz \quad \text{and} \quad RHS = 4xyz \]
Clearly, \( -6xyz \neq 4xyz \), which shows that the original statement is false.

### Conclusion:
Thus, we have shown that if \( x, y, z \) are in A.P., then the equation \( x^3 + z^3 - 8y^3 = 4xyz \) does not hold true.