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George Weigt 2006-05-04 09:09:33 -07:00
parent 516214639a
commit d699a9bdcd
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@ -50,7 +50,7 @@ $$x-1=-\root3\of2$$
$$(x-1)^3=-2$$
$$(x-1)(x^2-2x+1)=-2$$
$$x^3-2x^2+x-x^2+2x-1=-2$$
$$x^3-3x^3+3x-1=-2$$
$$x^3-3x^2+3x-1=-2$$
$$x^3-3x^2+3x+1=0$$
Let $p(x)=x^3-3x^2+3x+1$.
By the rational zeroes theorem the possible roots are $\pm1$.