Question

If \(\dfrac{a-b}{a+b} = \dfrac{(a-b)^3}{(a+b)^3}\), what set(s) of \(a\) and \(b\) is/are correct?

A. \(S_1 = \{a = \text{any real value}, b = 0\}\)
B. \(S_2 = \{a = 0, b = \text{any real value}\}\)
C. \(S_3 = \{a = 0, b = 0\}\)

• A. Only A
• B. Only B
• C. Only A and B
• D. All A, B and C

Hint:

Cross multiply and factor out (a-b)(a+b); the equation reduces to (a-b)(a+b)(4ab) = 0, so it holds whenever exactly one of a or b is zero, but not when both are zero at once, since that makes the original fraction 0/0.

Answer 1

The Correct Option is C

Step 1: Clear the denominators.
We are given \(\dfrac{a-b}{a+b} = \dfrac{(a-b)^3}{(a+b)^3}\), which only makes sense when \(a+b \neq 0\). Cross multiplying:
\[ (a-b)(a+b)^3 = (a-b)^3(a+b) \]

Step 2: Factor the equation.
Bring everything to one side and pull out the common factor \((a-b)(a+b)\):
\[ (a-b)(a+b)\left[(a+b)^2 - (a-b)^2\right] = 0 \]
Using the identity \((a+b)^2-(a-b)^2 = 4ab\), this becomes:
\[ (a-b)(a+b)(4ab) = 0 \]

Step 3: Read off the cases.
This product is zero when \(a = b\), or \(a = -b\), or \(a = 0\), or \(b = 0\). But we already fixed \(a + b \neq 0\), so the case \(a = -b\) is not allowed since it makes the original expression undefined. That leaves \(a = b\), \(a = 0\) (with \(b \neq 0\)), or \(b = 0\) (with \(a \neq 0\)) as the genuine solutions.

Step 4: Check each given set.
Set \(S_1\) (any a, b = 0): for \(a \neq 0\), \(\dfrac{a-0}{a+0} = 1\) and \(\dfrac{(a-0)^3}{(a+0)^3} = 1\). Both sides match, so \(S_1\) works.
Set \(S_2\) (a = 0, any b): for \(b \neq 0\), \(\dfrac{0-b}{0+b} = -1\) and \(\dfrac{(0-b)^3}{(0+b)^3} = -1\). Both sides match, so \(S_2\) also works.
Set \(S_3\) (a = 0, b = 0): the expression becomes \(\frac{0}{0}\), which is undefined. So \(S_3\) fails.

Final Answer:
Only \(S_1\) and \(S_2\) give a valid, satisfied equation. \(S_3\) breaks the expression entirely.
\[ \boxed{\text{Only A and B}} \]