23. Let $x$ and $y$ be positive real numbers such that $x^{3}+y^{3}+\frac{1}{27}=x y$. Find the value of $\frac{1}{x}$.
23. Ans: 3 .
Using the fact that $\mathrm{AM} \geq \mathrm{GM}$, we get $x^{3}+y^{3}+1 / 27 \geq 3 \sqrt[3]{x^{3} y^{3}(1 / 27)}=x y$ and $=$ holds only when $x^{3}=1 / 27$. Thus $x=1 / 3$.