Z treści zadania wiemy, że

$\text{tg}x+\text{ctg}x=3$  

Założenie: 

$x\ne \frac{k\pi }{2},k\in \mathbb{Z}$  

Zatem

• ${\text{tg}}^{2}x+{\text{ctg}}^{2}x=\text{(}\underset{=3}{\underbrace{\text{tg}x+\text{ctg}x}}{\text{)}}^2-2\text{tg}x\text{ctg}x=3^2-2\cdot \text{tg}x\cdot \frac{1}{\text{tg}x}=$

$=9-2\cdot 1=9-2=7$   

• ${\text{tg}}^{3}x+{\text{ctg}}^{3}x=\left(\text{tg}x+\text{ctg}x\right)\left({\text{tg}}^{2}x-\text{tg}x\text{ctg}x+{\text{ctg}}^{2}x\right)=3\left({\text{tg}}^{2}x-1+{\text{ctg}}^{2}x\right)=$  

$=3{\text{tg}}^{2}x-3+3{\text{ctg}}^{2}x=3\left({\text{tg}}^{2}x+{\text{ctg}}^{2}x\right)-3=3\left({\left(\text{tg}x+\text{ctg}x\right)}^2-2\text{tg}x\text{ctg}x\right)-3=$  

$=3\left(3^2-2\cdot 1\right)-3=3\left(9-2\right)-3=3\cdot 7-3=21-3=18$