How is this false?
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I start with x=4+23–√−−−−−−−√−3–√, then
x+3–√(x+3–√)2x2+(23–√)x+3x2+(23–√)x−1−23–√=4+23–√−−−−−−−√=(4+23–√−−−−−−−√)2=4+23–√=0
So I have shown that there is some polynomial whose solution is x=4+23–√−−−−−−−√−3–√, but I have not shown it to be 1.
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