mathschallenge.net

Problem

Without the use of a calculator determine which is greater in value, the square root of two or the cube root of three.

Solution

Let $x = \sqrt{2} = 2^{1/2}$ and $y = \sqrt[3]{3} = 3^{1/3}$.

$$\begin{align}\therefore x^6 &= \left(2^{1/2} \right)^6 = 2^3 = 8\\y^6 &= \left(3^{1/3} \right)^6 = 3^2 = 9\end{align}$$

As $y^6 \gt x^6$ it follows that $\sqrt[3]{3} \gt \sqrt{2}$.

Although this approach is perfectly valid and leads to the correct solution in this context, caution must be taken when negative numbers are involved. For example, from $(-2)^2 \gt 1^2$ it does not follow that $-2 \gt 1$.