Question Number 79015 by ajfour last updated on 22/Jan/20
![Rigorously over one month′s time, I developed a formula for general cubic. x^3 +ax^2 +bx+c=0 let x=((pt+q)/(t+1)) pq=m, p+q=s ________________________ m^2 {(a^2 +b)^2 −6a(ab−c)} +m{2(b^2 +ac)(a^2 +b)− 3(ab−c)(ab+3c)} +(b^2 +ac)^2 −6bc(ab−c)=0 ________________________ s=−(2/3){((m(a^2 +b)+b^2 +ac)/(ab−c))} +{(8/(27))[((m(a^2 +b)+b^2 +ac)/(ab−c))]^3 −8[((m^3 +bm^2 +acm+c^2 )/(ab−c))]}^(1/3) p,q = (s/2)±(√((s^2 /4)−m)) t=−(((3pq^2 +2apq+ap^2 +2bp+bq+3c))/((p^3 +ap^2 +bp+c))) x=((pt+q)/(t+1)) . (Please help checking..) (edited a digit 1 in place of 4)](https://www.tinkutara.com/question/Q79015.png)
$${Rigorously}\:{over}\:{one}\:{month}'{s} \\ $$$${time},\:{I}\:{developed}\:{a}\:{formula}\:{for} \\ $$$${general}\:{cubic}. \\ $$$${x}^{\mathrm{3}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${let}\:\:{x}=\frac{{pt}+{q}}{{t}+\mathrm{1}} \\ $$$$\boldsymbol{{pq}}=\boldsymbol{{m}},\:\boldsymbol{{p}}+\boldsymbol{{q}}=\boldsymbol{{s}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\boldsymbol{{m}}^{\mathrm{2}} \left\{\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)^{\mathrm{2}} −\mathrm{6}\boldsymbol{{a}}\left(\boldsymbol{{ab}}−\boldsymbol{{c}}\right)\right\} \\ $$$$+\boldsymbol{{m}}\left\{\mathrm{2}\left(\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}\right)\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)−\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left(\boldsymbol{{ab}}−\boldsymbol{{c}}\right)\left(\boldsymbol{{ab}}+\mathrm{3}\boldsymbol{{c}}\right)\right\} \\ $$$$\:\:\:+\left(\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}\right)^{\mathrm{2}} −\mathrm{6}\boldsymbol{{bc}}\left(\boldsymbol{{ab}}−\boldsymbol{{c}}\right)=\mathrm{0} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\boldsymbol{{s}}=−\frac{\mathrm{2}}{\mathrm{3}}\left\{\frac{\boldsymbol{{m}}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)+\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}}{\boldsymbol{{ab}}−\boldsymbol{{c}}}\right\} \\ $$$$\:\:+\left\{\frac{\mathrm{8}}{\mathrm{27}}\left[\frac{\boldsymbol{{m}}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}\right)+\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{ac}}}{\boldsymbol{{ab}}−\boldsymbol{{c}}}\right]^{\mathrm{3}} \right. \\ $$$$\left.\:\:\:−\mathrm{8}\left[\frac{\boldsymbol{{m}}^{\mathrm{3}} +\boldsymbol{{bm}}^{\mathrm{2}} +\boldsymbol{{acm}}+\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{ab}}−\boldsymbol{{c}}}\right]\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\boldsymbol{{p}},\boldsymbol{{q}}\:=\:\frac{\boldsymbol{{s}}}{\mathrm{2}}\pm\sqrt{\frac{\boldsymbol{{s}}^{\mathrm{2}} }{\mathrm{4}}−\boldsymbol{{m}}} \\ $$$$\boldsymbol{{t}}=−\frac{\left(\mathrm{3}\boldsymbol{{pq}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{apq}}+\boldsymbol{{ap}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{{bp}}+\boldsymbol{{bq}}+\mathrm{3}\boldsymbol{{c}}\right)}{\left(\boldsymbol{{p}}^{\mathrm{3}} +\boldsymbol{{ap}}^{\mathrm{2}} +\boldsymbol{{bp}}+\boldsymbol{{c}}\right)} \\ $$$$\boldsymbol{{x}}=\frac{\boldsymbol{{pt}}+\boldsymbol{{q}}}{\boldsymbol{{t}}+\mathrm{1}}\:. \\ $$$$\left({Please}\:{help}\:{checking}..\right) \\ $$$$\left({edited}\:{a}\:{digit}\:\mathrm{1}\:{in}\:{place}\:{of}\:\mathrm{4}\right) \\ $$
Commented by mr W last updated on 22/Jan/20
$${great}\:{sir}!\:{you}\:{did}\:{it}! \\ $$$${just}\:{one}\:{question}\:{before}\:{further} \\ $$$${studying}: \\ $$$${we}\:{get}\:{generally}\:{two}\:{values}\:{for}\:{m}\:{and} \\ $$$${s},\:{for}\:{each}\:{pair}\:{of}\:{m}\:{and}\:{s}\:{we}\:{get} \\ $$$${a}\:{pair}\:{of}\:{p}\:{and}\:{q}\:{and}\:{one}\:{value}\:{of}\:{t}, \\ $$$${and}\:{finally}\:{a}\:{root}\:{x}.\:{totally}\:{we}\:{get} \\ $$$${two}\:{roots}.\:{but}\:{generally}\:{we}\:{have}\:{three} \\ $$$${roots}.\:{what}\:{did}\:{i}\:{understand}\:{wrongly}? \\ $$
Commented by MJS last updated on 22/Jan/20
![doesn′t work for x^3 +3x^2 −13x−15=0 with x_1 =−5, x_2 =−1, x_3 =3 your method gives depending on the choice of p and q [t is not symmetric in p, q; so we have to decide p=(s/2)+... or p=(s/2)−...] x≈.172±.763 or x≈−1.43±2.53 btw. in this case m∉R](https://www.tinkutara.com/question/Q79037.png)
$$\mathrm{doesn}'\mathrm{t}\:\mathrm{work}\:\mathrm{for} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{13}{x}−\mathrm{15}=\mathrm{0} \\ $$$$\mathrm{with}\:{x}_{\mathrm{1}} =−\mathrm{5},\:{x}_{\mathrm{2}} =−\mathrm{1},\:{x}_{\mathrm{3}} =\mathrm{3} \\ $$$$\mathrm{your}\:\mathrm{method}\:\mathrm{gives}\:\mathrm{depending}\:\mathrm{on}\:\mathrm{the}\:\mathrm{choice} \\ $$$$\mathrm{of}\:{p}\:\mathrm{and}\:{q}\:\left[{t}\:\mathrm{is}\:\mathrm{not}\:\mathrm{symmetric}\:\mathrm{in}\:{p},\:{q};\:\mathrm{so}\:\mathrm{we}\right. \\ $$$$\left.\mathrm{have}\:\mathrm{to}\:\mathrm{decide}\:{p}=\frac{{s}}{\mathrm{2}}+…\:\mathrm{or}\:{p}=\frac{{s}}{\mathrm{2}}−…\right] \\ $$$${x}\approx.\mathrm{172}\pm.\mathrm{763}\:\mathrm{or}\:{x}\approx−\mathrm{1}.\mathrm{43}\pm\mathrm{2}.\mathrm{53} \\ $$$$\mathrm{btw}.\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:{m}\notin\mathbb{R} \\ $$
Commented by ajfour last updated on 22/Jan/20
$${its}\:{not}\:{at}\:{all}\:{good}!\:{but}\:{just}\:{now} \\ $$$${i}\:{had}\:{been}\:{modifying}\:{the} \\ $$$${Cardano}\:{formula},\:{first}\:{let}\:{me} \\ $$$${confirm}\:{and}\:{i}\:{shall}\:{post}\:{then}.. \\ $$
Commented by TawaTawa last updated on 23/Jan/20
$$\mathrm{Weldone}\:\mathrm{sir} \\ $$