6.089. $\left\{\begin{array}{l}u^{3}+v^{3}+1=m, \\ u^{3} v^{3}=-m .\end{array}\right.$
## Solution.

From the first equation of the system $v^{3}=m-1-u^{3}$. Substituting this value of $v^{3}$ into the second equation, we get

$$
u^{3}\left(m-1-u^{3}\right)=-m \Leftrightarrow\left(u^{3}\right)^{2}-(m-1) u^{3}-m=0,
$$

from which

$$
\begin{aligned}
& u_{1,2}^{3}=\frac{m-1 \pm \sqrt{(m-1)^{2}+4 m}}{2}=\frac{m-1 \pm \sqrt{m^{2}-2 m+1+4 m}}{2}= \\
& =\frac{m-1 \pm \sqrt{m^{2}+2 m+1}}{2}=\frac{m-1 \pm \sqrt{(m+1)^{2}}}{2}=\frac{m-1 \pm(m+1)}{2} \\
& u_{1}^{3}=\frac{m-1-m-1}{2}=-1, u_{1}=-1
\end{aligned}
$$

or

$$
u_{2}^{3}=\frac{m-1+m+1}{2}=m, u_{2}=\sqrt[3]{m}
$$

then $v_{1}^{3}=m, v_{1}=\sqrt[3]{m}$ or $v_{2}^{3}=m-1-m=-1, v_{2}=-1$.

Answer: $(-1 ; \sqrt[3]{m})(\sqrt[3]{m} ;-1)$