If `(1)/((ax+b)(cx+d))=(A)/(ax+b) + (B)/(cx+d)` then show that `(1)/((ax+b)^(2)(cx+d))=(A)/((ax+b)^(2))+(AB)/(ax+b)+(B^(2))/(cx+d)`.

Answer

To solve the problem step by step, we start with the given equation and manipulate it to reach the desired form.

### Step 1: Start with the given equation
We have the equation:
\[ \frac{1}{(ax + b)(cx + d)} = \frac{A}{ax + b} + \frac{B}{cx + d} \]

### Step 2: Combine the right-hand side
To combine the right-hand side, we need a common denominator:
\[ \frac{A}{ax + b} + \frac{B}{cx + d} = \frac{A(cx + d) + B(ax + b)}{(ax + b)(cx + d)} \]

### Step 3: Set the numerators equal
Since the denominators are equal, we can set the numerators equal to each other:
\[ 1 = A(cx + d) + B(ax + b) \]

### Step 4: Expand the right-hand side
Expanding the right-hand side gives:
\[ 1 = Acx + Ad + Bax + Bb \]
Rearranging this, we have:
\[ 1 = (Ac + Ba)x + (Ad + Bb) \]

### Step 5: Equate coefficients
For the equation to hold for all \(x\), the coefficients of \(x\) must be equal and the constant terms must be equal. Thus, we get:
1. \(Ac + Ba = 0\) (coefficient of \(x\))
2. \(Ad + Bb = 1\) (constant term)

### Step 6: Now, we need to show the second part
We need to show that:
\[ \frac{1}{(ax + b)^2(cx + d)} = \frac{A}{(ax + b)^2} + \frac{AB}{ax + b} + \frac{B^2}{cx + d} \]

### Step 7: Start with the left-hand side of the new equation
We start with:
\[ \frac{1}{(ax + b)^2(cx + d)} \]

### Step 8: Use the original equation
We can substitute the original equation into this expression:
\[ \frac{1}{(ax + b)^2(cx + d)} = \frac{1}{(ax + b)^2} \cdot \frac{1}{cx + d} \]
Using the original equation, we replace \(\frac{1}{cx + d}\):
\[ \frac{1}{(ax + b)^2(cx + d)} = \frac{1}{(ax + b)^2} \left( \frac{A}{ax + b} + \frac{B}{cx + d} \right) \]

### Step 9: Distribute the term
Distributing gives:
\[ \frac{1}{(ax + b)^2(cx + d)} = \frac{A}{(ax + b)^3} + \frac{B}{(ax + b)^2(cx + d)} \]

### Step 10: Combine and simplify
Now, we can combine the terms:
\[ \frac{A}{(ax + b)^2} + \frac{AB}{ax + b} + \frac{B^2}{cx + d} \]

### Conclusion
Thus, we have shown that:
\[ \frac{1}{(ax + b)^2(cx + d)} = \frac{A}{(ax + b)^2} + \frac{AB}{ax + b} + \frac{B^2}{cx + d} \]
Hence proved.