144. Extract the cube root $\sqrt[3]{2+\sqrt{5}}$.
Solution. We will look for rational numbers $a$ and $b$ such that $\sqrt[3]{2+\sqrt{5}}=a+b \sqrt{5}$.

Raising the obtained equality to the third power and equating the coefficients of $\sqrt{5}$ and the rational terms in both parts, we get the system:

$$
\left\{\begin{array}{l}
3 a^{2} b+5 b^{3}=1 \\
a^{3}+15 a b^{2}=2
\end{array}\right.
$$

Subtract the first equation, multiplied by 2, from the second equation, we get that $a^{3}-6 a^{2} b+15 a b^{2}-10 b^{3}=0$. Since $b \neq 0$, we have:

$$
\left(\frac{a}{b}\right)^{3}-6\left(\frac{a}{b}\right)^{2}+15\left(\frac{a}{b}\right)-10=0
$$

i.e., the number $\frac{a}{b}$ is a root of the equation $x^{3}-6 x^{2}+15 x-10=0$. It is easy to establish that one of the roots of this equation is 1, and it has no other real roots, since $x^{3}-6 x^{2}+$ $+15 x-10=(x-1)\left(x^{2}-5 x+10\right)$ and the discriminant of the quadratic $x^{2}-$ $-5 x+10$ is less than zero.

Thus, $\frac{a}{b}=1$, i.e., $a=b$, and from the system written above, we get that $a=b=\frac{1}{2}$.

Answer: $\sqrt[3]{2+\sqrt{5}}=\frac{1+\sqrt{5}}{2}$.