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Mean of three numbers is 0 then what is mean of their cubes

Sagot :

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]