If $x+y+z=0$ and $x=0.3$, $y=0.5$, $z=-0.8$, then what is $(x^3+y^3+z^3)\div(3xyz)$?
0
1
-1
3
Correct answer: 1
Solution
For three numbers with $x+y+z=0$, we have $x^3+y^3+z^3=3xyz$. Therefore, $(x^3+y^3+z^3)/(3xyz)=1$. The given values satisfy the condition, so the result is 1.