$a y^3 + b y^2 + c y + d = 0$

gdje su $a,b,c,d$ elementi u polju $F$ i $a\neq 0$.

Dakle, pišemo

$\array{ a y^3 + b y^2 + c y + d &=& a (y + \frac{b}{3 a})^3 - \frac{b^2}{3 a}y - \frac{b^3}{27 a^2} + c y + d\\ &=& a (y + \frac{b}{3 a})^3 + (c - \frac{b^2}{3 a})(y + \frac{b}{3 a}) + (d - \frac{b^3}{27 a^2} - \frac{ b c}{3 a} + \frac{b^3}{9 a^2}) \\ &=& a (y + \frac{b}{3 a})^3 + (c - \frac{b^2}{3 a})(y + \frac{b}{3 a}) + (d + \frac{2 b^3 - 9 a b c}{27 a^2}) \\ &=& a ( x^3 + p x + q), }$

gdje je $x = y + \frac{b}{3 a}$, $p = \frac{3 a c - b^2}{3 a}$ i $q = \frac{d}{a} + \frac{2 b^3 - 9 a b c}{27 a^3}$.

$x^3 + p x + q = 0.$

Tražimo $x$ u obliku (Vieteova supstitucija)

$x = w - \frac{p}{3 w}.$

Dakle,

$w^3 - p w + \frac{p^2}{3 w} + \frac{p^3}{27 w^3} + p w - \frac{p^2}{3 w} + q = 0$
$w^3 + \frac{p^3}{27 w^3} + q = 0$

Tu jednadžbu pomnožimo s $w^3$ i dobijemo kvadratnu jednadžbu za $w^3$,

$(w^3)^2 + q w^3 + \frac{p^3}{27} = 0,$

tako da bi trebalo biti

$w^3 = -\frac{q}{2} \pm\sqrt{\left(\frac{q}{2}\right)^2 - \frac{4 p^3}{27}}$

Created on April 4, 2019 at 09:56:04. See the history of this page for a list of all contributions to it.