If x^{2}+y^{2}+1=2x, then the value of x^{3}+ | vedclass

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(D) Given the equation: $x^{2}+y^{2}+1=2x$Rearranging the terms,we get: $x^{2}-2x+1+y^{2}=0$This can be written as: $(x-1)^{2}+y^{2}=0$Since the squar

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(D) Given the equation: $x^{2}+y^{2}+1=2x$Rearranging the terms,we get: $x^{2}-2x+1+y^{2}=0$This can be written as: $(x-1)^{2}+y^{2}=0$Since the square of any real number is non-negative,$(x-1)^{2} \geq 0$ and $y^{2} \geq 0$.For the sum of two non-negative numbers to be zero,each term must individually be zero:$(x-1)^{2}=0 \implies x=1$$y^{2}=0 \implies y=0$Substituting these values into the expression $x^{3}+y^{5}$:$x^{3}+y^{5} = (1)^{3} + (0)^{5} = 1 + 0 = 1$

If $x^{2}+y^{2}+1=2x,$ then the value of $x^{3}+y^{5}$ is

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