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

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(A) Given equation: $x^{2}+y^{2}+z^{2}+2=2(y-x)$Rearranging the terms: $x^{2}+2x+y^{2}-2y+z^{2}+2=0$Completing the squares: $(x^{2}+2x+1)+(y^{2}-2y+1)

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(A) Given equation: $x^{2}+y^{2}+z^{2}+2=2(y-x)$Rearranging the terms: $x^{2}+2x+y^{2}-2y+z^{2}+2=0$Completing the squares: $(x^{2}+2x+1)+(y^{2}-2y+1)+z^{2}=0$This simplifies to: $(x+1)^{2}+(y-1)^{2}+z^{2}=0$Since the sum of squares is zero,each term must be zero:$x+1=0 \Rightarrow x=-1$$y-1=0 \Rightarrow y=1$$z=0$Therefore,$x^{3}+y^{3}+z^{3} = (-1)^{3} + (1)^{3} + (0)^{3} = -1 + 1 + 0 = 0$.

If $x^{2}+y^{2}+z^{2}+2=2(y-x)$,then the value of $x^{3}+y^{3}+z^{3}$ is equal to

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