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 \( a^3 + 4b^3 + c^3 = 3b(a^2 + c^2) \) given that \( a, b, c \) are in Arithmetic Progression (A.P.), we can follow these steps:

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

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

### Step 3: Expand \( b^3 \)
Now, we expand \( \left(\frac{a+c}{2}\right)^3 \):
\[ \left(\frac{a+c}{2}\right)^3 = \frac{(a+c)^3}{8} \]
Using the binomial expansion:
\[ (a+c)^3 = a^3 + 3a^2c + 3ac^2 + c^3 \]
Thus,
\[ \left(\frac{a+c}{2}\right)^3 = \frac{a^3 + 3a^2c + 3ac^2 + c^3}{8} \]

### Step 4: Substitute back into LHS
Now substituting back into LHS:
\[ LHS = a^3 + 4\left(\frac{a^3 + 3a^2c + 3ac^2 + c^3}{8}\right) + c^3 \]
This simplifies to:
\[ LHS = a^3 + \frac{4(a^3 + 3a^2c + 3ac^2 + c^3)}{8} + c^3 \]
\[ = a^3 + \frac{1}{2}(a^3 + 3a^2c + 3ac^2 + c^3) + c^3 \]
\[ = a^3 + \frac{1}{2}a^3 + \frac{3}{2}a^2c + \frac{3}{2}ac^2 + \frac{1}{2}c^3 + c^3 \]
\[ = \frac{3}{2}a^3 + \frac{3}{2}a^2c + \frac{3}{2}ac^2 + \frac{3}{2}c^3 \]

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

### Step 6: Simplify the right-hand side (RHS)
Now, let’s 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
Now we need to show that:
\[ \frac{3}{2}(a^3 + a^2c + ac^2 + c^3) = \frac{3(a+c)(a^2 + c^2)}{2} \]
We can factor \( a^3 + a^2c + ac^2 + c^3 \) as:
\[ = (a+c)(a^2 + c^2) \]
Thus, we have:
\[ LHS = \frac{3}{2}(a+c)(a^2 + c^2) = RHS \]

### Conclusion
Therefore, we have proved that:
\[ a^3 + 4b^3 + c^3 = 3b(a^2 + c^2) \]