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Question Number 126363 by ajfour last updated on 19/Dec/20

x^3 −x−c=0 ; [c<2/(3(√3))] (Solve by a method other than trigonometric solution).

$${x}^{\mathrm{3}} −{x}−{c}=\mathrm{0}\:\:\:\:\:\:\:;\:\:\:\:\left[{c}<\mathrm{2}/\left(\mathrm{3}\sqrt{\mathrm{3}}\right)\right] \\ $$ $$\left({Solve}\:{by}\:{a}\:\:{method}\:{other}\:{than}\right. \\ $$ $$\left.{trigonometric}\:{solution}\right). \\ $$

Answered by mathmax by abdo last updated on 19/Dec/20

x^3 −x−c=0 let x=u+v so e⇒(u+v)^3 −(u+v)−c=0 ⇒ u^3 +v^3 +3uv(u+v)−(u+v)−c=0 ⇒ u^3 +v^3 −c +(3uv−1)(u+v)=0 ⇒ { ((u^3 +v^3 =c)),((3uv−1=0)) :} ⇒ { ((u^3 +v^3 =c ⇒ { ((u^3 +v^3 =c)),((u^3 v^3 =(1/(27)))) :})),((uv=(1/3))) :} so u^3 and v^3 are solution of t^2 −ct +(1/(27))=0 Δ=c^2 −(4/(27)) <0 due to c<(2/(3(√3))) ⇒t_1 =((c+i(√((4/(27))−c^2 )))/2) t_2 =((c−i(√((4/(27))−c^2 )))/2) ⇒x=^3 (√((c+i(√((4/(27))−c^2 )))/2)) +^3 (√((c−i(√((4/(27))−c^2 )))/2))

$$\mathrm{x}^{\mathrm{3}} −\mathrm{x}−\mathrm{c}=\mathrm{0}\:\mathrm{let}\:\mathrm{x}=\mathrm{u}+\mathrm{v}\:\:\mathrm{so}\:\mathrm{e}\Rightarrow\left(\mathrm{u}+\mathrm{v}\right)^{\mathrm{3}} −\left(\mathrm{u}+\mathrm{v}\right)−\mathrm{c}=\mathrm{0}\:\Rightarrow \\ $$ $$\mathrm{u}^{\mathrm{3}} \:+\mathrm{v}^{\mathrm{3}} \:+\mathrm{3uv}\left(\mathrm{u}+\mathrm{v}\right)−\left(\mathrm{u}+\mathrm{v}\right)−\mathrm{c}=\mathrm{0}\:\Rightarrow \\ $$ $$\mathrm{u}^{\mathrm{3}} \:+\mathrm{v}^{\mathrm{3}} \:−\mathrm{c}\:+\left(\mathrm{3uv}−\mathrm{1}\right)\left(\mathrm{u}+\mathrm{v}\right)=\mathrm{0}\:\Rightarrow\begin{cases}{\mathrm{u}^{\mathrm{3}} \:+\mathrm{v}^{\mathrm{3}} \:=\mathrm{c}}\\{\mathrm{3uv}−\mathrm{1}=\mathrm{0}}\end{cases} \\ $$ $$\Rightarrow\begin{cases}{\mathrm{u}^{\mathrm{3}} \:+\mathrm{v}^{\mathrm{3}} \:=\mathrm{c}\:\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\begin{cases}{\mathrm{u}^{\mathrm{3}} \:+\mathrm{v}^{\mathrm{3}} \:=\mathrm{c}}\\{\mathrm{u}^{\mathrm{3}} \mathrm{v}^{\mathrm{3}} \:=\frac{\mathrm{1}}{\mathrm{27}}}\end{cases}}\\{\mathrm{uv}=\frac{\mathrm{1}}{\mathrm{3}}}\end{cases} \\ $$ $$\mathrm{so}\:\mathrm{u}^{\mathrm{3}} \:\mathrm{and}\:\mathrm{v}^{\mathrm{3}} \:\mathrm{are}\:\mathrm{solution}\:\mathrm{of}\:\:\mathrm{t}^{\mathrm{2}} −\mathrm{ct}\:+\frac{\mathrm{1}}{\mathrm{27}}=\mathrm{0} \\ $$ $$\Delta=\mathrm{c}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{27}}\:<\mathrm{0}\:\mathrm{due}\:\mathrm{to}\:\:\:\mathrm{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:\Rightarrow\mathrm{t}_{\mathrm{1}} =\frac{\mathrm{c}+\mathrm{i}\sqrt{\frac{\mathrm{4}}{\mathrm{27}}−\mathrm{c}^{\mathrm{2}} }}{\mathrm{2}} \\ $$ $$\mathrm{t}_{\mathrm{2}} =\frac{\mathrm{c}−\mathrm{i}\sqrt{\frac{\mathrm{4}}{\mathrm{27}}−\mathrm{c}^{\mathrm{2}} }}{\mathrm{2}}\:\Rightarrow\mathrm{x}=^{\mathrm{3}} \sqrt{\frac{\mathrm{c}+\mathrm{i}\sqrt{\frac{\mathrm{4}}{\mathrm{27}}−\mathrm{c}^{\mathrm{2}} }}{\mathrm{2}}}\:+^{\mathrm{3}} \sqrt{\frac{\mathrm{c}−\mathrm{i}\sqrt{\frac{\mathrm{4}}{\mathrm{27}}−\mathrm{c}^{\mathrm{2}} }}{\mathrm{2}}} \\ $$

Commented byMJS_new last updated on 19/Dec/20

Cardano′s solution is not valid for D<0

$$\mathrm{Cardano}'\mathrm{s}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{not}\:\mathrm{valid}\:\mathrm{for}\:{D}<\mathrm{0} \\ $$

Commented bymathmax by abdo last updated on 20/Dec/20

why sir?

$$\mathrm{why}\:\mathrm{sir}? \\ $$

Commented byMJS_new last updated on 20/Dec/20

x^3 +px+q=0; D=(p^3 /(27))+(q^2 /4) if D<0 we have 3 real roots we cannot find using Cardano. see below article

$${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0};\:{D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}} \\ $$ $$\mathrm{if}\:{D}<\mathrm{0}\:\mathrm{we}\:\mathrm{have}\:\mathrm{3}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{find} \\ $$ $$\mathrm{using}\:\mathrm{Cardano}.\:\mathrm{see}\:\mathrm{below}\:\mathrm{article} \\ $$

Commented byMJS_new last updated on 20/Dec/20

https://en.m.wikipedia.org/wiki/Casus_irreducibilis

Commented byMJS_new last updated on 20/Dec/20

don′t get confused, in this article Δ is not equal to the D I use. still it′s true. you cannot reduce the complex number to its real value ⇒ the solution is possible to write down but impossible to use

$$\mathrm{don}'\mathrm{t}\:\mathrm{get}\:\mathrm{confused},\:\mathrm{in}\:\mathrm{this}\:\mathrm{article}\:\Delta\:\mathrm{is}\:{not} \\ $$ $$\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:{D}\:\mathrm{I}\:\mathrm{use}.\:\mathrm{still}\:\mathrm{it}'\mathrm{s}\:\mathrm{true}.\:\mathrm{you}\:\mathrm{cannot} \\ $$ $$\mathrm{reduce}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{number}\:\mathrm{to}\:\mathrm{its}\:\mathrm{real}\:\mathrm{value} \\ $$ $$\Rightarrow\:\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{write}\:\mathrm{down}\:\mathrm{but} \\ $$ $$\mathrm{impossible}\:\mathrm{to}\:\mathrm{use} \\ $$

Commented bymr W last updated on 20/Dec/20

that′s the point! certainly we can express the roots using cardano′s formula using complex numbers, but to calculate the real values out from the formula, we must at the end come to the trigonometric expression form. therefore cardano′s formula is not really useful in this case.

$${that}'{s}\:{the}\:{point}! \\ $$ $${certainly}\:{we}\:{can}\:{express}\:{the}\:{roots} \\ $$ $${using}\:{cardano}'{s}\:{formula}\:{using}\: \\ $$ $${complex}\:{numbers},\:{but}\:{to}\:{calculate} \\ $$ $${the}\:{real}\:{values}\:{out}\:{from}\:{the}\:{formula}, \\ $$ $${we}\:{must}\:{at}\:{the}\:{end}\:{come}\:{to}\:{the} \\ $$ $${trigonometric}\:{expression}\:{form}. \\ $$ $${therefore}\:{cardano}'{s}\:{formula}\:{is}\:{not} \\ $$ $${really}\:{useful}\:{in}\:{this}\:{case}. \\ $$

Commented bymathmax by abdo last updated on 20/Dec/20

nevermind thank you sir

$$\mathrm{nevermind}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented byTawa11 last updated on 23/Jul/21

great

$$\mathrm{great} \\ $$

Answered by ajfour last updated on 19/Dec/20

x=t+h ⇒ t^3 +3ht^2 +(3h^2 −1)t+h^3 −h−c=0 let t=p be also a root. ⇒ t^4 +(3h−p)t^3 +(3h^2 −1−3ph)t^2 +[h^3 −h−c−p(3h^2 −1)]t −p(h^3 −h−c)=0 lets add to it qt{t^3 +3ht^2 +(3h^2 −1)t+h^3 −h−c}=0 ⇒ (1+q)t^4 +(3h−p+3qh)t^3 + [3h^2 −1−3ph+q(3h^2 −1)]t^2 + [h^3 −h−c−p(3h^2 −1)+q(h^3 −h−c)]t −p(h^3 −h−c)=0 Now let 3ph=(1+q)(3h^2 −1);[coeff. of t^2 =0] ⇒ 3ht^4 +(6h^2 +1)t^3 + [3h(h^3 −h−c)−(3h^2 −1)^2 ]t −(3h^2 −1)(h^3 −h−c)=0 let h^3 +mh−c=0 ⇒ h^3 −h−c=−(m+1)h ⇒ 3ht^4 +(6h^2 +1)t^3 −[(3h^2 −1)^2 +3(m+1)h^2 ]t +(m+1)h(3h^2 −1)=0 ⇒ ......

$${x}={t}+{h}\:\:\Rightarrow \\ $$ $${t}^{\mathrm{3}} +\mathrm{3}{ht}^{\mathrm{2}} +\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){t}+{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$ $${let}\:\:{t}={p}\:\:{be}\:{also}\:{a}\:{root}.\:\:\:\Rightarrow \\ $$ $${t}^{\mathrm{4}} +\left(\mathrm{3}{h}−{p}\right){t}^{\mathrm{3}} +\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}−\mathrm{3}{ph}\right){t}^{\mathrm{2}} \\ $$ $$+\left[{h}^{\mathrm{3}} −{h}−{c}−{p}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)\right]{t} \\ $$ $$−{p}\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$ $${lets}\:{add}\:{to}\:{it} \\ $$ $${qt}\left\{{t}^{\mathrm{3}} +\mathrm{3}{ht}^{\mathrm{2}} +\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){t}+{h}^{\mathrm{3}} −{h}−{c}\right\}=\mathrm{0} \\ $$ $$\Rightarrow \\ $$ $$\left(\mathrm{1}+{q}\right){t}^{\mathrm{4}} +\left(\mathrm{3}{h}−{p}+\mathrm{3}{qh}\right){t}^{\mathrm{3}} + \\ $$ $$\left[\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}−\mathrm{3}{ph}+{q}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)\right]{t}^{\mathrm{2}} + \\ $$ $$\left[{h}^{\mathrm{3}} −{h}−{c}−{p}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)+{q}\left({h}^{\mathrm{3}} −{h}−{c}\right)\right]{t} \\ $$ $$−{p}\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$ $${Now}\:{let} \\ $$ $$\mathrm{3}{ph}=\left(\mathrm{1}+{q}\right)\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right);\left[{coeff}.\:{of}\:{t}^{\mathrm{2}} =\mathrm{0}\right] \\ $$ $$\Rightarrow\:\mathrm{3}{ht}^{\mathrm{4}} +\left(\mathrm{6}{h}^{\mathrm{2}} +\mathrm{1}\right){t}^{\mathrm{3}} + \\ $$ $$\left[\mathrm{3}{h}\left({h}^{\mathrm{3}} −{h}−{c}\right)−\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \right]{t} \\ $$ $$−\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$ $${let}\:\:\:{h}^{\mathrm{3}} +{mh}−{c}=\mathrm{0}\:\:\:\Rightarrow\: \\ $$ $$\:{h}^{\mathrm{3}} −{h}−{c}=−\left({m}+\mathrm{1}\right){h}\:\:\: \\ $$ $$\Rightarrow\:\:\:\:\:\:\mathrm{3}{ht}^{\mathrm{4}} +\left(\mathrm{6}{h}^{\mathrm{2}} +\mathrm{1}\right){t}^{\mathrm{3}} \\ $$ $$\:\:\:\:−\left[\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}\left({m}+\mathrm{1}\right){h}^{\mathrm{2}} \right]{t} \\ $$ $$\:\:\:\:+\left({m}+\mathrm{1}\right){h}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0}\:\:\:\:\Rightarrow \\ $$ $$...... \\ $$

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