6. Solve the system

$$
\left\{\begin{array}{l}
\operatorname{tg}^{3} x+\operatorname{tg}^{3} y+\operatorname{tg}^{3} z=36 \\
\operatorname{tg}^{2} x+\operatorname{tg}^{2} y+\operatorname{tg}^{2} z=14 \\
\left(\operatorname{tg}^{2} x+\operatorname{tg} y\right)(\operatorname{tg} x+\operatorname{tg} z)(\operatorname{tg} y+\operatorname{tg} z)=60
\end{array}\right.
$$

In the answer, indicate the sum of the minimum and maximum $\operatorname{tgx}$, which are solutions to the system.
6. Let $\operatorname{tg} x=x_{1}, \operatorname{tg} y=x_{2}, \operatorname{tg} z=x_{3}$.

Then $\left\{\begin{array}{l}x_{1}^{3}+x_{2}^{3}+x_{3}^{3}=36 \\ x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=14 \\ \left(x_{1}+x_{2}\right)\left(x_{1}+x_{3}\right)\left(x_{2}+x_{3}\right)=60\end{array}\right.$. Further, let $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=a \\ x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=b \text {. } \\ x_{1} x_{2} x_{3}=c\end{array}\right.$.

We will use the identities: $(x+y+z)^{3}-3(x+y+z)(x y+x z+y z)=x^{3}+y^{3}+z^{3}-3 x y z$

$$
x^{2}+y^{2}+z^{2}=(x+y+z)^{2}-2(x y+x z+y z)
$$

Then the system can be written as

$\left\{\begin{array}{l}a^{3}-3(a b-c)=36 \\ a^{2}-2 b=14 \\ a b-c=60\end{array} \Rightarrow\left\{\begin{array}{l}a^{3}-3 \cdot 60=36 \\ a^{2}-2 b=14 \\ a b-c=60\end{array} \Rightarrow a^{3}=216, a=6 \Rightarrow\right.\right.$

Then $2 b=36-14=22, b=11$. And finally $c=66-60=6$.

Thus, we have the system: $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=6 \\ x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}=11 \\ x_{1} x_{2} x_{3}=6\end{array}\right.$.

Consider the cubic equation in $x$ with roots $x_{1}, x_{2}, x_{3}$. Then by Vieta's theorem, we have $x^{3}-x^{2}\left(x_{1}+x_{2}+x_{3}\right)+x\left(x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}\right)-x_{1} x_{2} x_{3}=0$.

Further, we have: $x^{3}-6 x^{2}+11 x-6=0$. Let's find the roots of this equation. The first root will be obviously $x=1$, the other roots $x=2, x=3$.

Since the system is symmetric with respect to the cyclic permutation of the variables $x_{1}, x_{2}, x_{3}$, the system has six solutions, which are obtained by cyclic permutation of the numbers $1,2,3$. Therefore, the minimum value of $\operatorname{tgx}$ is 1, and the maximum is 3.

Thus, the answer is: $\{4\}$