Answer:
![(x^3+y^3+z^3)/(3) =0](https://img.qammunity.org/2022/formulas/mathematics/high-school/pzdfdukippvx8i4qewfhtk9gb9b6tzjfxj.png)
Explanation:
From the question we are told that
mean of three numbers is zero
Generally mean refers to average of number
Let
x, y, z be the three numbers with a mean of zero
T the mean of there cubes
Mathematically the mean of these three numbers is given as
![(x+y+z)/(3) =0](https://img.qammunity.org/2022/formulas/mathematics/high-school/rbztnw9uw3lhxq45zgm10eytom5rqkd7jl.png)
![x+y+z =0](https://img.qammunity.org/2022/formulas/mathematics/high-school/jjlqnh43eu5iq2kshjovyx74ttj2sozq2h.png)
and there cubes
![(x^3+y^3+z^3)/(3) =T](https://img.qammunity.org/2022/formulas/mathematics/high-school/bc1o758iwsic5ulysefbtdk6bka73gd043.png)
![x^3+y^3+z^3=3T](https://img.qammunity.org/2022/formulas/mathematics/high-school/4ax873ast36676eiciq3hccd55dohik4j3.png)
Mathematically solving the above equations by substitution method
![x+y+z =0......1](https://img.qammunity.org/2022/formulas/mathematics/high-school/cyzka6dkuhd2knew7f8ir2xa8jeytnmihs.png)
![x^3+y^3+z^3=3T ......2](https://img.qammunity.org/2022/formulas/mathematics/high-school/1d27cq51qiwjldfwvzzjlhtvsvt3heuo4i.png)
![x=-y-z ...3](https://img.qammunity.org/2022/formulas/mathematics/high-school/epgobclrpq2o5rjjwel8zeqc7cnpdyerxx.png)
equating 3 in 2
![-y^3-z^3+y^3+z^3=3T ..... 4\\3T=0](https://img.qammunity.org/2022/formulas/mathematics/high-school/1lnpag2unpck5v6a8teuvwihszrf5sy73c.png)
![T=0](https://img.qammunity.org/2022/formulas/mathematics/high-school/elwzdbhmncgz9h8xlcwgotuj7ep0vl3o0i.png)
Therefore the mean of the cubes of the three number is 0
![(x^3+y^3+z^3)/(3) =0](https://img.qammunity.org/2022/formulas/mathematics/high-school/pzdfdukippvx8i4qewfhtk9gb9b6tzjfxj.png)