If a,b,c are in A.P., then `a^(3)+c^(3)-8b^(3)` is equal to

Options

A. 2 abc
B. 3abc
C. 4abc
D. `-6 abc `

Explanation

To solve the problem, we need to find the value of \( a^3 + c^3 - 8b^3 \) given that \( a, b, c \) are in Arithmetic Progression (A.P.).

### Step-by-Step Solution:

1. **Understanding A.P.**:
Since \( a, b, c \) are in A.P., we can express \( b \) in terms of \( a \) and \( c \):
\[ b = \frac{a + c}{2} \]

2. **Expressing \( b^3 \)**:
We can cube \( b \):
\[ b^3 = \left(\frac{a + c}{2}\right)^3 = \frac{(a + c)^3}{8} \]

3. **Expanding \( (a + c)^3 \)**:
Using the binomial expansion:
\[ (a + c)^3 = a^3 + 3a^2c + 3ac^2 + c^3 \]
Therefore, we have:
\[ b^3 = \frac{a^3 + 3a^2c + 3ac^2 + c^3}{8} \]

4. **Substituting \( b^3 \) into the expression**:
Now we substitute \( b^3 \) into the expression \( a^3 + c^3 - 8b^3 \):
\[ a^3 + c^3 - 8b^3 = a^3 + c^3 - 8 \left(\frac{a^3 + 3a^2c + 3ac^2 + c^3}{8}\right) \]
Simplifying this gives:
\[ a^3 + c^3 - (a^3 + 3a^2c + 3ac^2 + c^3) \]
\[ = a^3 + c^3 - a^3 - c^3 - 3a^2c - 3ac^2 \]
\[ = -3a^2c - 3ac^2 \]

5. **Factoring out -3ac**:
We can factor out \(-3ac\):
\[ = -3ac(a + c) \]

6. **Substituting \( a + c \)**:
Since \( a + c = 2b \) (from the A.P. property), we can substitute:
\[ = -3ac(2b) = -6abc \]

### Final Result:
Thus, we find that:
\[ a^3 + c^3 - 8b^3 = -6abc \]