As all the fractions are raised to the power 3, the expression which has the largest value is that with the largest fraction in the brackets.

Each of these fractions is a little larger than $1\frac{1}{2}$. Subtracting $1\frac{1}{2}$ from each in turn, we get the fractions $\frac{1}{14}$, $\frac{1}{6}$, $\frac{1}{4}$, $\frac{3}{10}$, $0$, the largest of which is $\frac{3}{10}$ (because $0$ < $\frac{1}{14}$ < $\frac{1}{6}$ < $\frac{1}{4}$ = $\frac{2\frac{1}{2}}{10}$ < $\frac{3}{10}$).

Hence $\left(\frac{9}{5}\right)^3$ is the largest.

*This problem is taken from the UKMT Mathematical Challenges.*