Question Number 224819 by Abdulazim last updated on 05/Oct/25
Answered by fantastic last updated on 05/Oct/25
Answered by Ghisom_ last updated on 06/Oct/25
$$\left(\mathrm{2}{x}^{\mathrm{3}} +{x}−{a}\right)^{\mathrm{3}} ={a}−{x}^{\mathrm{3}} \\ $$$${a}=\mathrm{2}{x}^{\mathrm{3}} \\ $$$$\left(\mathrm{2}{x}^{\mathrm{3}} +{x}−\mathrm{2}{x}^{\mathrm{3}} \right)^{\mathrm{3}} =\mathrm{2}{x}^{\mathrm{3}} −{x}^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} ={x}^{\mathrm{3}} \:\mathrm{true} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{3}} =\frac{{a}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} =\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${x}_{\mathrm{2},\:\mathrm{3}} =\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} \mathrm{e}^{\pm\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\mathrm{but}\:\mathrm{there}\:\mathrm{are}\:\mathrm{3}\:\mathrm{more}\:\mathrm{conjugated}\:\mathrm{pairs}\:\mathrm{of} \\ $$$$\mathrm{complex}\:\mathrm{solutions}\:\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{only} \\ $$$$\mathrm{approximate} \\ $$
Commented by Ghisom_ last updated on 06/Oct/25
![in this case it′s possible to solve exactly but the exact solutions are not useful... (2x^3 +x−3)^3 =3−x^3 the trick: let x^3 =a (x+2a−3)^3 =3−a x^3 +3(2a−3)x^2 +3(2a−3)^2 x+(8a^3 −36a^2 +55a−30=0 let x^3 =a 3(2a−3)(x^2 +(2a−3)x+((2(2a^2 −6a+5))/3))=0 a=(3/2) ⇒ x^3 =(3/2) [see above] from now on a≠(3/2) x^2 +(2a−3)x+((2(2a^2 −6a+5))/3)=0 (A) obviously x≠0 so we can multiply by x x^3 +(2a−3)x^2 +((2(2a^2 −6a+5))/3)x=0 let x^3 =a & transform x^2 +((3(2a^2 −6a+5))/(3(2a−3)))x+(a/(2a−3))=0 (B) now subtract (A)−(B) and solve for x x=−((8a^3 −36a^2 +53a−30)/(8a^2 −24a+17)) insert in (A) and transform a^6 −9a^5 +((69)/2)a^4 −72a^3 +((5493)/(64))a^2 −((3519)/(64))a+((125)/8)=0 let a=t+(3/2) t^6 +(3/4)t^4 −((15)/(64))t^2 +((247)/(256))=0 let t=±(√u) u^3 +(3/4)u^2 −((15)/(64))u+((247)/(256))=0 let u=v−(1/4) v^3 −((27)/(64))v+((135)/(128))=0 we can solve this, we get 3 values for v inserting backwards we get a=((3±(√(4v−1)))/2)=x^3 [we get 6 values for a] x=a^(1/3) (−(1/2)+((√3)/2)i)^k with k=0, 1, 2 now we have 18 values for x but we introduced false solutions, so we have to check all of them to get the 6 true ones...](https://www.tinkutara.com/question/Q224830.png)
$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{it}'\mathrm{s}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{exactly} \\ $$$$\mathrm{but}\:\mathrm{the}\:\mathrm{exact}\:\mathrm{solutions}\:\mathrm{are}\:\mathrm{not}\:\mathrm{useful}… \\ $$$$\left(\mathrm{2}{x}^{\mathrm{3}} +{x}−\mathrm{3}\right)^{\mathrm{3}} =\mathrm{3}−{x}^{\mathrm{3}} \\ $$$$\mathrm{the}\:\mathrm{trick}:\:\mathrm{let}\:{x}^{\mathrm{3}} ={a} \\ $$$$\left({x}+\mathrm{2}{a}−\mathrm{3}\right)^{\mathrm{3}} =\mathrm{3}−{a} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}\left(\mathrm{2}{a}−\mathrm{3}\right){x}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{2}{a}−\mathrm{3}\right)^{\mathrm{2}} {x}+\left(\mathrm{8}{a}^{\mathrm{3}} −\mathrm{36}{a}^{\mathrm{2}} +\mathrm{55}{a}−\mathrm{30}=\mathrm{0}\right. \\ $$$$\mathrm{let}\:{x}^{\mathrm{3}} ={a} \\ $$$$\mathrm{3}\left(\mathrm{2}{a}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\left(\mathrm{2}{a}−\mathrm{3}\right){x}+\frac{\mathrm{2}\left(\mathrm{2}{a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{5}\right)}{\mathrm{3}}\right)=\mathrm{0} \\ $$$${a}=\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:{x}^{\mathrm{3}} =\frac{\mathrm{3}}{\mathrm{2}}\:\left[\mathrm{see}\:\mathrm{above}\right] \\ $$$$\mathrm{from}\:\mathrm{now}\:\mathrm{on}\:{a}\neq\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\left(\mathrm{2}{a}−\mathrm{3}\right){x}+\frac{\mathrm{2}\left(\mathrm{2}{a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{5}\right)}{\mathrm{3}}=\mathrm{0}\:\:\:\:\:\left({A}\right) \\ $$$$\mathrm{obviously}\:{x}\neq\mathrm{0}\:\mathrm{so}\:\mathrm{we}\:\mathrm{can}\:\mathrm{multiply}\:\mathrm{by}\:{x} \\ $$$${x}^{\mathrm{3}} +\left(\mathrm{2}{a}−\mathrm{3}\right){x}^{\mathrm{2}} +\frac{\mathrm{2}\left(\mathrm{2}{a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{5}\right)}{\mathrm{3}}{x}=\mathrm{0} \\ $$$$\mathrm{let}\:{x}^{\mathrm{3}} ={a}\:\&\:\mathrm{transform} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{3}\left(\mathrm{2}{a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{5}\right)}{\mathrm{3}\left(\mathrm{2}{a}−\mathrm{3}\right)}{x}+\frac{{a}}{\mathrm{2}{a}−\mathrm{3}}=\mathrm{0}\:\:\:\:\:\left({B}\right) \\ $$$$\mathrm{now}\:\mathrm{subtract}\:\left({A}\right)−\left({B}\right)\:\mathrm{and}\:\mathrm{solve}\:\mathrm{for}\:{x} \\ $$$${x}=−\frac{\mathrm{8}{a}^{\mathrm{3}} −\mathrm{36}{a}^{\mathrm{2}} +\mathrm{53}{a}−\mathrm{30}}{\mathrm{8}{a}^{\mathrm{2}} −\mathrm{24}{a}+\mathrm{17}} \\ $$$$\mathrm{insert}\:\mathrm{in}\:\left({A}\right)\:\mathrm{and}\:\mathrm{transform} \\ $$$${a}^{\mathrm{6}} −\mathrm{9}{a}^{\mathrm{5}} +\frac{\mathrm{69}}{\mathrm{2}}{a}^{\mathrm{4}} −\mathrm{72}{a}^{\mathrm{3}} +\frac{\mathrm{5493}}{\mathrm{64}}{a}^{\mathrm{2}} −\frac{\mathrm{3519}}{\mathrm{64}}{a}+\frac{\mathrm{125}}{\mathrm{8}}=\mathrm{0} \\ $$$$\mathrm{let}\:{a}={t}+\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${t}^{\mathrm{6}} +\frac{\mathrm{3}}{\mathrm{4}}{t}^{\mathrm{4}} −\frac{\mathrm{15}}{\mathrm{64}}{t}^{\mathrm{2}} +\frac{\mathrm{247}}{\mathrm{256}}=\mathrm{0} \\ $$$$\mathrm{let}\:{t}=\pm\sqrt{{u}} \\ $$$${u}^{\mathrm{3}} +\frac{\mathrm{3}}{\mathrm{4}}{u}^{\mathrm{2}} −\frac{\mathrm{15}}{\mathrm{64}}{u}+\frac{\mathrm{247}}{\mathrm{256}}=\mathrm{0} \\ $$$$\mathrm{let}\:{u}={v}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${v}^{\mathrm{3}} −\frac{\mathrm{27}}{\mathrm{64}}{v}+\frac{\mathrm{135}}{\mathrm{128}}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this},\:\mathrm{we}\:\mathrm{get}\:\mathrm{3}\:\mathrm{values}\:\mathrm{for}\:{v} \\ $$$$\mathrm{inserting}\:\mathrm{backwards}\:\mathrm{we}\:\mathrm{get} \\ $$$${a}=\frac{\mathrm{3}\pm\sqrt{\mathrm{4}{v}−\mathrm{1}}}{\mathrm{2}}={x}^{\mathrm{3}} \:\:\:\:\:\left[\mathrm{we}\:\mathrm{get}\:\mathrm{6}\:\mathrm{values}\:\mathrm{for}\:{a}\right] \\ $$$${x}={a}^{\mathrm{1}/\mathrm{3}} \left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)^{{k}} \:\mathrm{with}\:{k}=\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{18}\:\mathrm{values}\:\mathrm{for}\:{x}\:\mathrm{but}\:\mathrm{we} \\ $$$$\mathrm{introduced}\:\mathrm{false}\:\mathrm{solutions},\:\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{to}\:\mathrm{check}\:\mathrm{all}\:\mathrm{of}\:\mathrm{them}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{6}\:\mathrm{true}\:\mathrm{ones}… \\ $$
Commented by behi834171 last updated on 06/Oct/25
$${nice}\:{solution}\:{sir}.{thank}\:{you}\:{so}\:{much} \\ $$$${for}\:{beautiful}\:{methods}. \\ $$$${sir}!\:{are}\:{you}\:{former}\:{member}\:{of}\:{forum} \\ $$$${that}\:{i}\:{have}\:{him}\:{in}\:{my}\:{mind}\:{as}: \\ $$$${MJS}\:{sir}\:? \\ $$$${or},{not}? \\ $$
Commented by Ghisom_ last updated on 07/Oct/25
$$\mathrm{some}\:\mathrm{people}\:\mathrm{think}\:\mathrm{MJS}\:=\:\mathrm{Frix} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{into}\:\mathrm{any}\:\mathrm{of}\:\mathrm{these}\:\mathrm{games} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{just}\:\mathrm{myself}\:\mathrm{and}\:\mathrm{I}\:\mathrm{love}\:\mathrm{mathematics} \\ $$
Commented by Frix last updated on 08/Oct/25
$$\mathrm{Well}\:\mathrm{well}\:\mathrm{well} \\ $$
Commented by necx122 last updated on 08/Oct/25
$${I}'{ve}\:{always}\:{wanted}\:{to}\:{ask}\:{this}\:{question}. \\ $$$${Ghisom}\:{types}\:{and}\:{solves}\:{like}\:{MJS} \\ $$$${back}\:{in}\:{the}\:{days}. \\ $$