If a, b, c are in A.P., prove that `a^(3)+4b^(3)+c^(3)=3b(a^(2)+c^(2)).`

Explanation

To prove that if \( a, b, c \) are in A.P., then \( a^3 + 4b^3 + c^3 = 3b(a^2 + c^2) \), we can follow these steps:

### Step 1: Use the property of A.P.
Since \( a, b, c \) are in Arithmetic Progression (A.P.), we have:
\[ b = \frac{a + c}{2} \]

### Step 2: Substitute \( b \) in the left-hand side (LHS)
We need to prove:
\[ a^3 + 4b^3 + c^3 = 3b(a^2 + c^2) \]
Substituting \( b \) into the LHS:
\[ LHS = a^3 + 4\left(\frac{a + c}{2}\right)^3 + c^3 \]

### Step 3: Expand \( b^3 \)
Calculating \( \left(\frac{a + c}{2}\right)^3 \):
\[ \left(\frac{a + c}{2}\right)^3 = \frac{(a + c)^3}{8} \]
Using the binomial expansion:
\[ (a + c)^3 = a^3 + 3a^2c + 3ac^2 + c^3 \]
Thus:
\[ \left(\frac{a + c}{2}\right)^3 = \frac{a^3 + 3a^2c + 3ac^2 + c^3}{8} \]
Now substituting back into the LHS:
\[ LHS = a^3 + 4 \cdot \frac{a^3 + 3a^2c + 3ac^2 + c^3}{8} + c^3 \]
\[ = a^3 + \frac{4(a^3 + 3a^2c + 3ac^2 + c^3)}{8} + c^3 \]
\[ = a^3 + \frac{a^3 + 3a^2c + 3ac^2 + c^3}{2} + c^3 \]

### Step 4: Combine like terms
Now combine the terms:
\[ = a^3 + c^3 + \frac{a^3 + 3a^2c + 3ac^2 + c^3}{2} \]
\[ = \frac{2a^3 + 2c^3 + a^3 + 3a^2c + 3ac^2 + c^3}{2} \]
\[ = \frac{3a^3 + 3c^3 + 3a^2c + 3ac^2}{2} \]
\[ = \frac{3(a^3 + c^3 + a^2c + ac^2)}{2} \]

### Step 5: Factor out common terms
Factor out \( 3 \):
\[ = \frac{3}{2}(a^3 + c^3 + ac(a + c)) \]

### Step 6: Simplify the right-hand side (RHS)
Now we simplify the right-hand side (RHS):
\[ RHS = 3b(a^2 + c^2) = 3\left(\frac{a + c}{2}\right)(a^2 + c^2) \]
\[ = \frac{3(a + c)(a^2 + c^2)}{2} \]

### Step 7: Show LHS = RHS
We need to show:
\[ \frac{3}{2}(a^3 + c^3 + ac(a + c)) = \frac{3(a + c)(a^2 + c^2)}{2} \]
This can be verified by expanding \( (a + c)(a^2 + c^2) \):
\[ = a^3 + ac^2 + c^3 + a^2c \]
Thus, both sides are equal, proving that:
\[ a^3 + 4b^3 + c^3 = 3b(a^2 + c^2) \]

### Conclusion
Hence, we have proved that if \( a, b, c \) are in A.P., then:
\[ a^3 + 4b^3 + c^3 = 3b(a^2 + c^2) \]