Which of the following is the greatest?

### Concept & Strategy To compare different surds (roots) of different numbers, you cannot easily judge them in their radical form. The most reliable strategy is to convert them to fractional exponents and then raise all the terms to a common power. This common power should be the Least Common Multiple (LCM) of the denominators of the fractional exponents. $$x^{\frac{1}{a}} \text{ and } y^{\frac{1}{b}} \rightarrow \text{Raise to LCM}(a,b)$$ ### Step-by-Step Solution * **Given:** The numbers are $\sqrt{2}, \sqrt[3]{3}, \sqrt[4]{4}, \sqrt[6]{6}$. * **Calculation / Deduction:** 1. Convert the radicals into fractional exponents: $$2^{\frac{1}{2}}, 3^{\frac{1}{3}}, 4^{\frac{1}{4}}, 6^{\frac{1}{6}}$$ 2. Identify the denominators of the exponents: 2, 3, 4, and 6. 3. Find the LCM of these denominators: $$\text{LCM}(2, 3, 4, 6) = 12$$ 4. Raise each term to the power of 12 to eliminate the fractions, making them easy to compare: * $(2^{\frac{1}{2}})^{12} = 2^{12/2} = 2^6 = 64$ * $(3^{\frac{1}{3}})^{12} = 3^{12/3} = 3^4 = 81$ * $(4^{\frac{1}{4}})^{12} = 4^{12/4} = 4^3 = 64$ * $(6^{\frac{1}{6}})^{12} = 6^{12/6} = 6^2 = 36$ 5. Compare the resulting integers: $81 > 64 = 64 > 36$. 6. The greatest value is 81, which corresponds to the original term $\sqrt[3]{3}$. ### Exam Strategy & Shortcut **Quick Simplification Check:** Notice immediately that $\sqrt[4]{4} = (2^2)^{\frac{1}{4}} = 2^{\frac{2}{4}} = 2^{\frac{1}{2}} = \sqrt{2}$. Since options (a) and (c) are mathematically identical, neither can be the unique "greatest" answer in a standard MCQ. This immediately narrows your choices down to just comparing $\sqrt[3]{3}$ and $\sqrt[6]{6}$. Raise both to the power of 6: $(\sqrt[3]{3})^6 = 3^2 = 9$ $(\sqrt[6]{6})^6 = 6^1 = 6$ Since $9 > 6$, $\sqrt[3]{3}$ is clearly the greatest. ### Common Pitfall Students often attempt to approximate decimal values for these roots without calculating them properly, leading to guesses. Without memorizing $\sqrt[3]{3} \approx 1.442$, guessing based on the base numbers (thinking 6 is the largest base so it must be greatest) is a very common trap. ### Final Answer **Therefore, the correct answer is $\sqrt[3]{3}$.**