If x^{2}+y^{2}+z^{2}=2(x+z-1),then find the v | vedclass

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(D) Given the equation: $x^{2}+y^{2}+z^{2}=2(x+z-1)$.Expanding the right side: $x^{2}+y^{2}+z^{2}=2x+2z-2$.Rearranging the terms to one side: $x^{2}-2

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$2$

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(D) Given the equation: $x^{2}+y^{2}+z^{2}=2(x+z-1)$.Expanding the right side: $x^{2}+y^{2}+z^{2}=2x+2z-2$.Rearranging the terms to one side: $x^{2}-2x+y^{2}+z^{2}-2z+2=0$.We can rewrite this as: $(x^{2}-2x+1) + y^{2} + (z^{2}-2z+1) = 0$.This simplifies to: $(x-1)^{2} + y^{2} + (z-1)^{2} = 0$.Since the sum of squares of real numbers is zero only if each term is zero,we have:$x-1=0 \Rightarrow x=1$$y=0$$z-1=0 \Rightarrow z=1$Now,calculate $x^{3}+y^{3}+z^{3} = (1)^{3} + (0)^{3} + (1)^{3} = 1 + 0 + 1 = 2$.

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

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