Choose z such that [itex]yz+1=0[/itex], or [itex]z=-{1 \over y}[/itex]
Then: [itex]y^3 - {1 \over y^3} + 6 = 0[/itex]
[itex]\Rightarrow (y^3)^2 + 6(y^3) - 1 = 0[/itex]

Solve as a quadratic equation and back substituting z gives:
[tex]x=y+z=\sqrt[3]{-3 + \sqrt{10}} - {1 \over \sqrt[3]{-3 + \sqrt{10}}}[/tex]
or
[tex]x=y+z=-\sqrt[3]{3 + \sqrt{10}} + {1 \over \sqrt[3]{3 + \sqrt{10}}}[/tex]

TBH, I learned it from an old book of my father, that he had while studying.
I don't recall what it was called, but it was a thick volume with a purple cover and many yellowed pages... I loved it!