一元三次方程求根公式详细逐步推导

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ax3+bx2+cx+d=0a0a,b,c,d

x=?

x3+bx2a+cxa+da=0

k0=1,k1=ba,k2=ca,k3=da

k0x3+k1x2+k2x+k3=0

x=y+η

k0(y+η)3+k1(y+η)2+k2(y+η)+k3=0

(y+η)3(y+η)2=y3+3y2η+3yη2+η3=y2+2yη+η2

k0y3+k03y2η+3yη2k0+η3k0+y2k1+2yηk1+η2k1+yk2+ηk2+k3=0

y3k0+y2(k03η+k1)+y1(3η2k0+2ηk1+k2)+y0(η3k0+η2k1+ηk2+k3)=0

k0=3k0η+k1=3η2k0+2ηk1+k2=η3k0+η2k1+ηk2+k3=10pq

y3+py+q=0

3η+k1=3η2+2ηk1+k2=η3+η2k1+ηk2+k3=0pq

η=k13p=3(k13)2+2(k13)k1+k2p=(k1)23+k2q=(k13)3+(k13)2k1+(k13)k2+k3q=k3127+3k3127k1k23+k3q=2k3127k1k23+k3

y=A13+B13A=A13A0,A1=ωA0,A2=ω2A0B=B13B0,B1=ωB0,B2=ω2B0

y3=A+B+(AB)13(A13+B13)y3=A+B+(AB)13yy3(A+B)(AB)13y=0y3+py+q=0p=3(AB)13q=(A+B)A+B=qAB=(p3)3


QuadraticFormula:x1+x2=bax1x2=caax2+bx+c=0A=x1,B=x2,q=ba,(3p)3=caax2+bx+c=0x1=b+b24ac2ax2=bb24ac2ax1=b+b24ac2a=b2a+b222a2cax2=b+b24ac2a=b2ab222a2ca


A0=q2+q24+p327B0=q2q24+p327A=A13A0,A1=ωA0,A2=ω2A0B=B13B0,B1=ωB0,B2=ω2B0ω=1±3i2AB=(q3)3qA0=(A0)3B0=(B0)3A1=(A1)3B1=(B2)3A2=(A2)3B2=(B1)3y=A13+B13y0=A0+B0y1=A1+B2y2=A2+B1y0=q2+(q24+p327)1/23+q2(q24+p327)1/23y1=1+3i2q2+(q24+p327)1/23+13i2q2(q24+p327)1/23y2=13i2q2+(q24+p327)1/23+1+3i2q2(q24+p327)1/23x=y+ηη=k13k1=bay=x+b3ap=(k1)23+k2q=2k3127k1k23+k3

x0+b3a=k3127+k1k26k32+(k3127k1k26d2a)2+(k2k213)3273+k3127+k1k26k3214(2k3127k1k23+k3)2+(k2k213)3273x1+b3a=1+3i2k3127+k1k26k32+14(2k3127k1k23+k3)2+(k2k213)3273+13i2k3127+k1k26k3214(2k3127k1k23+k3)2+(k2k213)3273x2+b3a=13i2k3127+k1k26k32+14(2k3127k1k23+k3)2+(k2k213)3273+1+3i2k3127+k1k26k3214(2k3127k1k23+k3)2+(k2k213)3273

k1=ba,k2=ca,k3=da
x0=b3a+b327a3+bc6a2d2a+14(2d327a3bc3a2+da)2+(c3ab29a2)33+b327a3+bc6a2d2a14(2d327a3bc3a2+da)2+(c3ab29a2)33x1=b3a+1+3i2b327a3+bc6a2d2a+14(2d327a3bc3a2+da)2+(c3ab29a2)33+13i2b327a3+bc6a2d2a14(2d327a3bc3a2+da)2+(c3ab29a2)33x2=b3a+13i2b327a3+bc6a2d2a+14(2d327a3bc3a2+da)2+(c3ab29a2)33+1+3i2b327a3+bc6a2d2a14(2d327a3bc3a2+da)2+(c3ab29a2)33

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