//注:颜色为代数置换标记,推导后期需将前面的设定值回代//注:颜色为代数置换标记,推导后期需将前面的设定值回代//注:颜色为代数置换标记,推导后期需将前面的设定值回代
\quad
已知:ax3+bx2+cx+d=0a≠0a,b,c,d∈ℜ问题:x=?
\begin{aligned}
&已知:ax^3+bx^2+cx+d=0 \qquad a\neq0 \qquad a,b,c,d\in\Re \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \\
&问题:\mathbf x\mathbf=\mathbf?
\end{aligned}
已知:ax3+bx2+cx+d=0a=0a,b,c,d∈ℜ问题:x=?
\quad
推导开始:推导开始:推导开始:
\quad
x3+bx2a+cxa+da=0a≠0a,b,c,d∈ℜ
\begin{aligned}
&x^3+\dfrac{bx^2}{a}+\dfrac{cx}{a}+\dfrac{d}{a}=0 \qquad a\neq0 \qquad a,b,c,d\in\Re \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad
\end{aligned}
x3+abx2+acx+ad=0a=0a,b,c,d∈ℜ
设k0=1,k1=ba,k2=ca,k3=dak0x3+k1x2+k2x+k3=0设x=y+ηk0(y+η)3+k1(y+η)2+k2(y+η)+k3=0(y+η)3=y3+3y2η+3yη2+η3(y+η)2=y2+2yη+η2k0y3+k03y2η+3yη2k0+η3k0+y2k1+2yηk1+η2k1+yk2+ηk2+k3=0y3k0+y2(k03η+k1)+y1(3η2k0+2ηk1+k2)+y0(η3k0+η2k1+ηk2+k3)=0
\begin{aligned}
&\color{red}{设\quad k_0=1,k_1=\dfrac{b}{a},k_2=\dfrac{c}{a},k_3=\dfrac{d}{a}}\\
&k_0x^3+k_1x^2+k_2x+k_3=0\\
\\
&\color{orange}{设 \quad x=y+\eta}\\
&k_0(y+\eta)^3+k_1(y+\eta)^2+k_2(y+\eta)+k_3=0\\
\\
&(y+\eta)^3=y^3+3y^2\eta+3y\eta^2+\eta^3\\
&(y+\eta)^2=y^2+2y\eta+\eta^2\\
\\
&k_0y^3+k_03y^2\eta+3y\eta^2k_0+\eta^3k_0+y^2k_1+2y\eta k_1+\eta^2k_1+yk_2+\eta k_2+k_3=0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\\
&y^3k_0+y^2(k_03\eta+k_1)+y^1(3\eta^2k_0+2\eta k_1+k_2)+y^0(\eta^3k_0+\eta^2k_1+\eta k_2+k_3)=0
\\
\end{aligned}
设k0=1,k1=ab,k2=ac,k3=adk0x3+k1x2+k2x+k3=0设x=y+ηk0(y+η)3+k1(y+η)2+k2(y+η)+k3=0(y+η)3=y3+3y2η+3yη2+η3(y+η)2=y2+2yη+η2k0y3+k03y2η+3yη2k0+η3k0+y2k1+2yηk1+η2k1+yk2+ηk2+k3=0y3k0+y2(k03η+k1)+y1(3η2k0+2ηk1+k2)+y0(η3k0+η2k1+ηk2+k3)=0
k0=13k0η+k1=03η2k0+2ηk1+k2=pη3k0+η2k1+ηk2+k3=q \begin{aligned} k_0=&1 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ 3k_0\eta+k_1=&0\\ 3\eta^2k_0+2\eta k_1+k_2=&p\\ \eta^3k_0+\eta^2k_1+\eta k_2+k_3=&q \end{aligned} k0=3k0η+k1=3η2k0+2ηk1+k2=η3k0+η2k1+ηk2+k3=10pq
设y3+py+q=03η+k1=03η2+2ηk1+k2=pη3+η2k1+ηk2+k3=q \begin{aligned} 设 \quad y^3+py+q=&0 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ 3\eta+k_1=&0\\ 3\eta^2+2\eta k_1+k_2=&p\\ \eta^3+\eta^2k_1+\eta k_2+k_3=&q\\ \end{aligned} 设y3+py+q=3η+k1=3η2+2ηk1+k2=η3+η2k1+ηk2+k3=00pq
η=−k13p=3(−k13)2+2(−k13)k1+k2p=−(k1)23+k2q=(−k13)3+(−k13)2k1+(−k13)k2+k3q=−k1327+3k1327−k1k23+k3q=2k1327−k1k23+k3 \begin{aligned} &\color{orange}{\eta=-\dfrac{k_1}{3}}\\ &p=3\left(-\dfrac{k_1}{3}\right)^2+2\left(-\dfrac{k_1}{3}\right)k_1+k_2\\ &\color{green}{p=-\dfrac{(k_1)^2}{3}+k_2}\\ &q=\left(-\dfrac{k_1}{3}\right)^3+\left(-\dfrac{k_1}{3}\right)^2k_1+\left(-\dfrac{k_1}{3}\right)k_2+k_3 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\\ &q=-\dfrac{k_1^3}{27}+\dfrac{3k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3\\ &\color{green}{q=\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3}\\ \end{aligned} η=−3k1p=3(−3k1)2+2(−3k1)k1+k2p=−3(k1)2+k2q=(−3k1)3+(−3k1)2k1+(−3k1)k2+k3q=−27k13+273k13−3k1k2+k3q=272k13−3k1k2+k3
设y=A13+B13A′=A13A0′∈ℜ,A1′=ωA0′,A2′=ω2A0′B′=B13B0′∈ℜ,B1′=ωB0′,B2′=ω2B0′∵y3=A+B+(AB)13(A13+B13)y3=A+B+(AB)13yy3−(A+B)−(AB)13y=0y3+py+q=0 \begin{aligned} 设 &\color{pink}{y=A^\frac13+B^\frac13\qquad A'=A^\frac13\quad A'_0\in\Re,A'_1=\omega A'_0,A'_2=\omega^2 A'_0 \quad B'=B^\frac13 \quad B'_0\in\Re,B'_1=\omega B'_0,B'_2=\omega^2 B'_0}\\ \because &y^3=A+B+(AB)^\frac13(A^\frac13+B^\frac13) \quad \quad \quad \quad \quad \quad \quad\\ &y^3=A+B+(AB)^\frac13y\\ &y^3-(A+B)-(AB)^\frac13y=0\\ &y^3+py+q=0\\ \end{aligned} 设∵y=A31+B31A′=A31A0′∈ℜ,A1′=ωA0′,A2′=ω2A0′B′=B31B0′∈ℜ,B1′=ωB0′,B2′=ω2B0′y3=A+B+(AB)31(A31+B31)y3=A+B+(AB)31yy3−(A+B)−(AB)31y=0y3+py+q=0
∴p=−3(AB)13q=−(A+B)A+B=−qAB=(−p3)3 \begin{aligned} \therefore &p=-3(AB)^\frac13 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\\ &q=-(A+B)\\ &A+B=-q \qquad AB=(-\frac{p}{3})^3 \\ \end{aligned} ∴p=−3(AB)31q=−(A+B)A+B=−qAB=(−3p)3
//分界线中的x,a,b,c为展示定理用,与上下文中的x,a,b,c无关根据韦达定理:∵x1+x2=−bax1x2=caax2+bx+c=0∴A=x1,B=x2,q=ba,(−p3)3=ca根据QuadraticFormula:∵ax2+bx+c=0x1=−b+b2−4ac2ax2=−b−b2−4ac2ax1=−b+b2−4ac2a=−b2a+b222a2−cax2=−b+b2−4ac2a=−b2a−b222a2−ca \begin{aligned} &//分界线中的x,a,b,c为展示定理用,与上下文中的x,a,b,c无关 \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\\ \\ &根据韦达定理:\\ &\because x_1+x_2=-\frac{b}{a} \qquad x_1x_2=\frac{c}{a}\qquad ax^2+bx+c=0\\ &\therefore A=x_1,B=x_2,q=\frac{b}{a},(-\frac{p}{3})^3=\frac{c}{a}\\ &根据Quadratic Formula: \\ & \because ax^2+bx+c=0 \\ & x_1=\frac{-b+\sqrt{b^2-4ac}}{2a} \qquad x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}\\ &x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}=-\frac{b}{2a}+\sqrt{\frac{b^2}{2^2a^2}-\frac{c}{a}}\\ &x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}=-\frac{b}{2a}-\sqrt{\frac{b^2}{2^2a^2}-\frac{c}{a}}\\ \end{aligned} //分界线中的x,a,b,c为展示定理用,与上下文中的x,a,b,c无关根据韦达定理:∵x1+x2=−abx1x2=acax2+bx+c=0∴A=x1,B=x2,q=ab,(−3p)3=ac根据QuadraticFormula:∵ax2+bx+c=0x1=2a−b+b2−4acx2=2a−b−b2−4acx1=2a−b+b2−4ac=−2ab+22a2b2−acx2=2a−b+b2−4ac=−2ab−22a2b2−ac
∴A0=−q2+q24+p327B0=−q2−q24+p327∵A′=A13A0′∈ℜ,A1′=ωA0′,A2′=ω2A0′B′=B13B0′∈ℜ,B1′=ωB0′,B2′=ω2B0′∵ω=−1±3i2AB=(−q3)3q∈ℜ∴A0=(A0′)3B0=(B0′)3A1=(A1′)3B1=(B2′)3A2=(A2′)3B2=(B1′)3∵y=A13+B13∴y0=A0′+B0′y1=A1′+B2′y2=A2′+B1′ \begin{aligned} &\therefore A_0=-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}\qquad B_0=-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}\\ &\because \color{pink}{A'=A^\frac13\quad A'_0\in\Re,A'_1=\omega A'_0,A'_2=\omega^2 A'_0 \quad B'=B^\frac13 \quad B'_0\in\Re,B'_1=\omega B'_0,B'_2=\omega^2 B'_0}\quad \quad \quad \quad \quad \quad \quad \quad\\ &\because \omega=\dfrac{-1\pm\sqrt{3}i}{2} \quad AB=(-\frac{q}{3})^3 \quad q\in\Re\\ &\therefore A_0=(A'_0)^3 \quad B_0=(B'_0)^3\\ &\quad A_1=(A'_1)^3 \quad B_1=(B'_2)^3\\ &\quad A_2=(A'_2)^3 \quad B_2=(B'_1)^3\\ &\because y=A^\frac13+B^\frac13 \quad \\ &\therefore y_0=A'_0+B'_0\\ &\quad y_1=A'_1+B'_2\\ &\quad y_2=A'_2+B'_1\\ \end{aligned} ∴A0=−2q+4q2+27p3B0=−2q−4q2+27p3∵A′=A31A0′∈ℜ,A1′=ωA0′,A2′=ω2A0′B′=B31B0′∈ℜ,B1′=ωB0′,B2′=ω2B0′∵ω=2−1±3iAB=(−3q)3q∈ℜ∴A0=(A0′)3B0=(B0′)3A1=(A1′)3B1=(B2′)3A2=(A2′)3B2=(B1′)3∵y=A31+B31∴y0=A0′+B0′y1=A1′+B2′y2=A2′+B1′
y0=−q2+(q24+p327)123+−q2−(q24+p327)123y1=−1+3i2−q2+(q24+p327)123+−1−3i2−q2−(q24+p327)123y2=−1−3i2−q2+(q24+p327)123+−1+3i2−q2−(q24+p327)123 \begin{aligned} \\ \quad y_0=\sqrt[3]{-\frac{q}{2}+(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}+\sqrt[3]{-\frac{q}{2}-(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}\quad \quad \quad \quad \quad \quad \quad \quad \quad\\ \\ \quad y_1=\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\frac{q}{2}+(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}+\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\frac{q}{2}-(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}\\ \\ \quad y_2=\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\frac{q}{2}+(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}+\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\frac{q}{2}-(\frac{q^2}{4}+\frac{p^3}{27})^{\frac{1}{2}}}\\ \\ \end{aligned} y0=3−2q+(4q2+27p3)21+3−2q−(4q2+27p3)21y1=2−1+3i3−2q+(4q2+27p3)21+2−1−3i3−2q−(4q2+27p3)21y2=2−1−3i3−2q+(4q2+27p3)21+2−1+3i3−2q−(4q2+27p3)21
∵x=y+η\because \color{orange}{x=y+\eta}∵x=y+η
η=−k13\quad \color{orange}{\eta=-\dfrac{k_1}{3}}η=−3k1
k1=ba\color{red}{\quad k_1=\dfrac{b}{a}}k1=ab
∴y=x+b3a\therefore y=x+\frac{b}{3a}∴y=x+3ab
∵p=−(k1)23+k2\because \color{green}{p=-\dfrac{(k_1)^2}{3}+k_2}∵p=−3(k1)2+k2
q=2k1327−k1k23+k3\quad \color{green}{q=\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3}q=272k13−3k1k2+k3
∴x0+b3a=−k1327+k1k26−k32+14(k1327−k1k26−d2a)2+(k2−k123)3273+−k1327+k1k26−k32−14(2k1327−k1k23+k3)2+(k2−k123)3273x1+b3a=−1+3i2−k1327+k1k26−k32+14(2k1327−k1k23+k3)2+(k2−k123)3273+−1−3i2−k1327+k1k26−k32−14(2k1327−k1k23+k3)2+(k2−k123)3273x2+b3a=−1−3i2−k1327+k1k26−k32+14(2k1327−k1k23+k3)2+(k2−k123)3273+−1+3i2−k1327+k1k26−k32−14(2k1327−k1k23+k3)2+(k2−k123)3273 \begin{aligned} \\ \therefore \quad x_0+\frac{b}{3a}=\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}+\sqrt{\frac{1}{4}(\dfrac{k_1^3}{27}-\dfrac{k_1k_2}{6}-\frac{d}{2a})^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}&\\ \qquad+\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}-\sqrt{\frac{1}{4}(\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3)^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}\\ \\ \quad x_1+\frac{b}{3a}=\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}+\sqrt{\frac{1}{4}(\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3)^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}&\\ \qquad+\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}-\sqrt{\frac{1}{4}(\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3)^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}\\ \\ \quad x_2+\frac{b}{3a}=\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}+\sqrt{\frac{1}{4}(\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3)^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}&\\ \qquad+\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\dfrac{k_1^3}{27}+\dfrac{k_1k_2}{6}-\frac{k_3}{2}-\sqrt{\frac{1}{4}(\dfrac{2k_1^3}{27}-\dfrac{k_1k_2}{3}+k_3)^2+\frac{(k_2-\frac{k_1^2}{3})^3}{27}}}\\ \\ \end{aligned} ∴x0+3ab=3−27k13+6k1k2−2k3+41(27k13−6k1k2−2ad)2+27(k2−3k12)3+3−27k13+6k1k2−2k3−41(272k13−3k1k2+k3)2+27(k2−3k12)3x1+3ab=2−1+3i3−27k13+6k1k2−2k3+41(272k13−3k1k2+k3)2+27(k2−3k12)3+2−1−3i3−27k13+6k1k2−2k3−41(272k13−3k1k2+k3)2+27(k2−3k12)3x2+3ab=2−1−3i3−27k13+6k1k2−2k3+41(272k13−3k1k2+k3)2+27(k2−3k12)3+2−1+3i3−27k13+6k1k2−2k3−41(272k13−3k1k2+k3)2+27(k2−3k12)3
∵k1=ba,k2=ca,k3=da\because \color{red}{k_1=\dfrac{b}{a},k_2=\dfrac{c}{a},k_3=\dfrac{d}{a}}∵k1=ab,k2=ac,k3=ad
∴x0=−b3a+−b327a3+bc6a2−d2a+14(2d327a3−bc3a2+da)2+(c3a−b29a2)33+−b327a3+bc6a2−d2a−14(2d327a3−bc3a2+da)2+(c3a−b29a2)33x1=−b3a+−1+3i2−b327a3+bc6a2−d2a+14(2d327a3−bc3a2+da)2+(c3a−b29a2)33+−1−3i2−b327a3+bc6a2−d2a−14(2d327a3−bc3a2+da)2+(c3a−b29a2)33x2=−b3a+−1−3i2−b327a3+bc6a2−d2a+14(2d327a3−bc3a2+da)2+(c3a−b29a2)33+−1+3i2−b327a3+bc6a2−d2a−14(2d327a3−bc3a2+da)2+(c3a−b29a2)33 \\ \\ \begin{aligned} \\ \\ \therefore \quad x_0=-\frac{b}{3a}+\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}+\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}&\\ +\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}-\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}\\ \\ \\ \quad x_1=-\frac{b}{3a}+\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}+\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}&\\ +\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}-\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}\\ \\ \\ \quad x_2=-\frac{b}{3a}+\dfrac{-1-\sqrt{3}i}{2}\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}+\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}&\\ +\dfrac{-1+\sqrt{3}i}{2}\sqrt[3]{-\dfrac{b^3}{27a^3}+\dfrac{bc}{6a^2}-\frac{d}{2a}-\sqrt{\frac{1}{4}(\dfrac{2d^3}{27a^3}-\dfrac{bc}{3a^2}+\frac{d}{a})^2+(\frac{c}{3a}-\frac{b^2}{9a^2})^3}}\\ \end{aligned} ∴x0=−3ab+3−27a3b3+6a2bc−2ad+41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3+3−27a3b3+6a2bc−2ad−41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3x1=−3ab+2−1+3i3−27a3b3+6a2bc−2ad+41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3+2−1−3i3−27a3b3+6a2bc−2ad−41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3x2=−3ab+2−1−3i3−27a3b3+6a2bc−2ad+41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3+2−1+3i3−27a3b3+6a2bc−2ad−41(27a32d3−3a2bc+ad)2+(3ac−9a2b2)3
附:一元三次方程的故事
塔尔塔利亚是意大利人,出生于1500年。他12岁那年,被入侵的法国兵砍伤了头部和舌头,从此说话结结巴巴,人们就给他一个绰号“塔尔塔利亚”(在意大利语中,这是口吃的意思),真名反倒少有人叫了,他自学成才,成了数学家,宣布自己找到了三次方程的的解法。这时,意大利数学家卡丹出场,请求塔尔塔利把解方程的方法告诉他,可是遭到了拒绝。后来卡丹对塔尔塔利假装说要推荐他去当西班牙炮兵顾问,称自己因为无法解三次方程而内心痛苦并发誓永远不泄漏塔尔塔利亚解一元三次方程式的秘密。塔尔塔利亚这才把解一元三次方程的秘密告诉了卡丹。六年以后,卡丹不顾原来的信约,在他的著作《关于代数的大法》中,将经过改进的三次方程的解法公开发表。后人就把这个方法叫作卡丹公式,塔尔塔利亚的名字反而被湮没了,正如他的真名在口吃以后被埋没了一样。
塔尔塔利亚对卡丹的背信行为非常恼怒,互相写信指骂对方。最终在一个不明的夜晚,卡丹派人秘密刺杀了塔尔塔利亚。
一元三次方程应有三个根。塔塔利亚公式给出的只是一个实根。又过了大约200年后,随着人们对虚数认识的加深,到了1732年,才由瑞士数学家欧拉找到了一元三次方程三个根的完整的表达式。
维护日志:
2019-12-16:review
2020-10-24:排版调整