Answer:
[tex]\frac{x^3+y^3+z^3}{3} =0[/tex]
Step-by-step 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
[tex]\frac{x+y+z}{3} =0[/tex]
[tex]x+y+z =0[/tex]
and there cubes
[tex]\frac{x^3+y^3+z^3}{3} =T[/tex]
[tex]x^3+y^3+z^3=3T[/tex]
Mathematically solving the above equations by substitution method
[tex]x+y+z =0......1[/tex]
[tex]x^3+y^3+z^3=3T ......2[/tex]
[tex]x=-y-z ...3[/tex]
equating 3 in 2
[tex]-y^3-z^3+y^3+z^3=3T ..... 4\\3T=0[/tex]
[tex]T=0[/tex]
Therefore the mean of the cubes of the three number is 0
[tex]\frac{x^3+y^3+z^3}{3} =0[/tex]
