8. An integer $x$ satisfies the inequality $x^{2} \leq 729 \leq-x^{3}$. $P$ and $Q$ are possible values of $x$. What is the maximum possible value of $10(P-Q)$ ?
Solution
180

We observe that $729=3^{6}$. First consider $x^{2} \leq 729$. This has solution $-27 \leq x \leq 27$. Now consider $729 \leq-x^{3}$. This may be rearranged to give $x^{3} \leq-729$ with solution $x \leq-9$.

These inequalities are simultaneously satisfied when $-27 \leq x \leq-9$. The maximum value of $P-Q$ is therefore $-9-(-27)=18$. So the answer to the question is $10 \times 18=180$.