B3. For a number $x$ it holds that $x+\frac{1}{x}=5$. We define $n=x^{3}+\frac{1}{x^{3}}$. It turns out that $n$ is an integer. Calculate $n$. (Write $n$ in ordinary decimal notation.)
B3. 110 For $x$ it holds that $x+\frac{1}{x}=5$, so $x^{2}-5 x+1=0$. From this it follows that $x=x_{1,2}=\frac{5 \pm \sqrt{21}}{2}$. Note that $x_{1} x_{2}=1$. Working out gives $x^{3}+x^{-3}=x_{1}^{3}+x_{2}^{3}=\frac{1}{8}\left((5+\sqrt{21})^{3}+(5-\sqrt{21})^{3}\right)=$ $\frac{1}{8}\left(\left(5^{3}+3 \cdot 5^{2} \cdot \sqrt{21}+3 \cdot 5 \cdot \sqrt{21}^{2}+\sqrt{21}^{3}\right)+\left(5^{3}-3 \cdot 5^{2} \cdot \sqrt{21}+3 \cdot 5 \cdot \sqrt{21}^{2}-\sqrt{21}^{3}\right)\right)=$ $\frac{2}{8}\left(5^{3}+3 \cdot 5 \cdot 21\right)=110$.

Alternative solution: From $\left(x+\frac{1}{x}\right)^{3}=x^{3}+3 x^{2}\left(\frac{1}{x}\right)+3 x\left(\frac{1}{x}\right)^{2}+\left(\frac{1}{x}\right)^{3}=x^{3}+3 x+\frac{3}{x}+\frac{1}{x^{3}}$ it follows that $x^{3}+\frac{1}{x^{3}}=\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)=5^{3}-3 \times 5=110$.