1.at3+bt2+ct+d=0⇒{\displaystyle at^{3}+bt^{2}+ct+d=0\Rightarrow } 以 t=x−b3a{\displaystyle t=x-{\frac {b}{3a}}} 代入
2.得 x3+px+q=0{\displaystyle x^{3}+px+q=0} ,令其三根為 x1,x2,x3{\displaystyle x_{1},x_{2},x_{3}}
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5.故 y1,y2{\displaystyle y_{1},y_{2}} 為 y2+qy+−(p3)3=0{\displaystyle y^{2}+qy+-({\frac {p}{3}})^{3}=0} 的兩根
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題目:x3−3x+2=0{\displaystyle x^{3}-3x+2=0}
題目:x3−12x−16=0{\displaystyle x^{3}-12x-16=0}