168. Find the number which, when multiplied by $3+\sqrt{5}$, gives one.
168. Al-Karaji believes:

\[
\begin{gathered}
x(3+\sqrt{5})=1 ; 3 x+x \sqrt{5}=1 ; 3 x+\sqrt{5 x^{2}}=1 \\
\sqrt{5 x^{2}}=1-3 x ; 5 x^{2}=(1-3 x)^{2}=1-6 x+9 x^{2} \\
4 x^{2}-6 x+1=0 ; x=\frac{3 \pm \sqrt{5}}{4}
\end{gathered}
\]

Al-Karaji takes the root \( x_{2}=\frac{3-\sqrt{5}}{4} \) as the desired one \((x_{1}>1)\). The usual method:

\[
x=\frac{1}{3+\sqrt{5}}=\frac{3-\sqrt{5}}{(3-\sqrt{5})(3+\sqrt{5})}=\frac{3-\sqrt{5}}{4}
\]