If a, b, c are in geometric progression, then prove that :
`(1)/(a^(2)-b^(2))+(1)/(b^(2))=(1)/(b^(2)-c^(2))`

Explanation

To prove that

\[ \frac{1}{a^2 - b^2} + \frac{1}{b^2} = \frac{1}{b^2 - c^2} \]

given that \(a\), \(b\), and \(c\) are in geometric progression, we can start by using the property of geometric progression.

### Step 1: Use the property of geometric progression
Since \(a\), \(b\), and \(c\) are in geometric progression, we have:

\[ b^2 = ac \]

### Step 2: Rewrite the left-hand side (LHS)
We start with the left-hand side:

\[ \text{LHS} = \frac{1}{a^2 - b^2} + \frac{1}{b^2} \]

### Step 3: Substitute \(b^2\) in terms of \(a\) and \(c\)
Using the relation \(b^2 = ac\), we can rewrite \(a^2 - b^2\):

\[ a^2 - b^2 = a^2 - ac = a(a - c) \]

Now substituting this back into the LHS:

\[ \text{LHS} = \frac{1}{a(a - c)} + \frac{1}{ac} \]

### Step 4: Find a common denominator
The common denominator for the two fractions is \(ac(a - c)\):

\[ \text{LHS} = \frac{c}{ac(a - c)} + \frac{(a - c)}{ac(a - c)} \]

### Step 5: Combine the fractions
Now we can combine the fractions:

\[ \text{LHS} = \frac{c + (a - c)}{ac(a - c)} = \frac{a}{ac(a - c)} \]

### Step 6: Simplify the expression
Now simplify the expression:

\[ \text{LHS} = \frac{1}{c(a - c)} \]

### Step 7: Rewrite the right-hand side (RHS)
Now we will rewrite the right-hand side:

\[ \text{RHS} = \frac{1}{b^2 - c^2} \]

Using \(b^2 = ac\):

\[ b^2 - c^2 = ac - c^2 = c(a - c) \]

Thus, the RHS becomes:

\[ \text{RHS} = \frac{1}{c(a - c)} \]

### Step 8: Compare LHS and RHS
Now we see that:

\[ \text{LHS} = \frac{1}{c(a - c)} = \text{RHS} \]

Thus, we have proved that:

\[ \frac{1}{a^2 - b^2} + \frac{1}{b^2} = \frac{1}{b^2 - c^2} \]

### Conclusion
Hence, the statement is proven.

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