Question

If $\frac{2^{\frac{1}{3}} + 27a^{\frac{5}{3}}}{(\frac{a}{3})^{-\frac{1}{3}} + 3a^{\frac{2}{3}}} = 2 - ka^{\frac{1}{3}} + 9a^{\frac{2}{3}}$, then $k = $?

• A. $\sqrt{2}$
• B. $\sqrt{6}$
• C. $3\sqrt{2}$
• D. $3\sqrt{6}$

Hint:

Use the identity $a^3+b^3 = (a+b)(a^2-ab+b^2)$ to simplify.

Answer 1

The Correct Option is D

Step 1: Concept
Simplify the algebraic expression using exponent rules.

Step 2: Analysis
The denominator $(\frac{a}{3})^{-\frac{1}{3}} + 3a^{\frac{2}{3}}$ can be rewritten as $3^{\frac{1}{3}}a^{-\frac{1}{3}} + 3a^{\frac{2}{3}}$. Multiply both numerator and denominator by $a^{\frac{1}{3}}$ to clear negative exponents. Numerator: $2^{\frac{1}{3}}a^{\frac{1}{3}} + 27a^{2} = (2a)^{\frac{1}{3}} + (3a^{\frac{2}{3}})^{3}$. This follows the form $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$.

Step 3: Calculation
By comparing both sides after simplification, the term $-ka^{\frac{1}{3}}$ corresponds to $-3\sqrt[3]{2} \cdot \sqrt[3]{27} a^{\frac{1}{3}}$, leading to $k = 3\sqrt{6}$.

Step 4: Conclusion
Hence, $k = 3\sqrt{6}$. Final Answer: (D)