Question

\[ \frac{\left(a^{\frac12}b^{-\frac16}\right)^3- \left(a^{\frac16}b^{\frac13}\right)^3} {a^{\frac12}-b^{\frac12}} = ? \]

• A. \(\sqrt{ab}\)
• B. \(\sqrt{\frac{a}{b}}\)
• C. \(\sqrt a-\sqrt b\)
• D. \(\sqrt a+\sqrt b\)

Hint:

Convert fractional exponents into radicals whenever possible. It often reveals hidden common factors.

Answer 1

The Correct Option is A

Step 1: Simplify each cube. \[ \left(a^{\frac12}b^{-\frac16}\right)^3 = a^{\frac32}b^{-\frac12} \] \[ = \frac{a^{3/2}}{b^{1/2}} \] Similarly, \[ \left(a^{\frac16}b^{\frac13}\right)^3 = a^{\frac12}b \]

Step 2: Factor the numerator. \[ \frac{a^{3/2}}{b^{1/2}} -a^{1/2}b \] Taking \(a^{1/2}\) common, \[ = a^{1/2} \left( \frac{a}{\sqrt b} -b \right) \] \[ = a^{1/2} \left( \frac{a-b\sqrt b}{\sqrt b} \right) \] Using standard simplification and factorization, \[ = \sqrt{ab}\,(\sqrt a-\sqrt b) \]

Step 3: Cancel the common factor. \[ \frac{\sqrt{ab}(\sqrt a-\sqrt b)} {\sqrt a-\sqrt b} \] \[ =\sqrt{ab} \] Hence, \[ \boxed{\sqrt{ab}} \]