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

Instead of factoring the general equation, we can test each set with real numbers and see which ones actually make \(\dfrac{a-b}{a+b}\) equal to \(\dfrac{(a-b)^3}{(a+b)^3}\).

1. Set S1 (a = any real value, b = 0): try a = 5, b = 0. Left side: \(\dfrac{5-0}{5+0} = 1\). Right side: \(\dfrac{5^3}{5^3} = 1\). Both sides equal 1, so this set satisfies the equation for any nonzero a.
2. Set S2 (a = 0, b = any real value): try a = 0, b = 5. Left side: \(\dfrac{0-5}{0+5} = -1\). Right side: \(\dfrac{(-5)^3}{5^3} = \dfrac{-125}{125} = -1\). Both sides equal -1, so this set also works for any nonzero b.
3. Set S3 (a = 0, b = 0): here both the top and bottom of the original fraction become 0, since a - b = 0 and a + b = 0. A fraction with a zero denominator has no value, so the equation cannot hold. This set fails.

So testing confirms that S1 and S2 always give matching, well defined values on both sides, while S3 breaks the expression before we can even compare the two sides.

Let's summarize:

• Whenever exactly one of a, b is zero and the other is not, both sides of the equation reduce to the same value, either 1 or -1.
• When both a and b are zero together, the expression is undefined, so that case cannot be called a solution.

The correct choice is Only A and B.