If $$x^{2} + y^{2} + 1 = 2x$$, then the value of $$x^{3} + y^{5}$$ is
$$x^{2} + y^{2} + 1 = 2x$$
$$x^{2} -2x + y^{2} + 1 = 0$$
$$(x^{2} -2x + 1 )+ y^{2} = 0$$
$$(x-1)^{2}+ y^{2} = 0$$
Since $$(x-1)^{2}$$ ≥ 0 for all x, and $$y^{2}$$ ≥ 0 for all y
and $$(x-1)^{2}+ y^{2} = 0$$, then
$$(x-1)^{2}$$ = 0 and $$y^{2}$$ = 0
So, x = 1, y = 0
Hence,
$$x^{3}+y^{5} = 1$$
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