Question 1043698
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x^2+y^2+z^2=2(x+z-1) then the value of x^3+y^3+z^3
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If  {{{x^2+y^2+z^2}}} = {{{2(x+z-1)}}},  then

{{{x^2 - 2x + y^2 + z^2 - 2z}}} = {{{-2}}},

{{{(x-1)^2 - 1 + y^2 + (z-1)^2 - 1}}} = {{{-2}}},

{{{(x-1)^2 + y^2 + (z-1)^2}}} = {{{0}}}.

We have the sum of squares of three real numbers equal to zero.
Hence, each of the three numbers is equal to zero:

x-1 = 0,  y = 0  and  z-1 = 0.

In other words, x = 1, y = 0, z = 1.

Now {{{x^3 + y^3 + z^3}} = {{{(-1)^3 + 0^3 + (-1)^3}}} = -2.
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Solved.