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p-3-q-3-c-pq-1-3-0-lt-c-lt-2-3-3-Find-p-q-

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Question Number 187388 by ajfour last updated on 16/Feb/23
p^3 +q^3 =c , pq=(1/3) ∀ 0<c<(2/(3(√3))) Find p+q.
$${p}^{\mathrm{3}} +{q}^{\mathrm{3}} ={c}\:\:\:,\:\:{pq}=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\forall\:\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$${Find}\:{p}+{q}. \\ $$
Answered by anurup last updated on 17/Feb/23
(p+q)^3 −3pq(p+q)=c x^3 −x−c=0,x=p+q x=u+v ⇒x^3 =u^3 +v^3 +3uvx ⇒x^3 −3uvx−(u^3 +v^3 )=0 ∴uv=(1/3) and u^3 +v^3 =c (u^3 −v^3 )^2 =(u^3 +v^3 )^2 −4u^3 v^3 ⇒(u^3 −v^3 )=±(√(c^2 −(4/(27)))) ∴u^3 =(1/2)(c±(√(c^2 −(4/(27)))) ),v^3 =(1/2)(c∓(√(c^2 −(4/(27))))) ∵u^3 v^3 =(1/(27)), u^3 =(1/2)(c+(√(c^2 −(4/(27)))) ),v^3 =(1/2)(c−(√(c^2 −(4/(27)))) ) ∵0<c<(2/(3(√3))), c^2 <(4/(27)) ∴u^3 =(1/2)(c+i(√((4/(27))−c^2 )) ),v^3 =(1/2)(c−i(√((4/(27))−c^2 )) ) u=(1/( (√3)))(cos θ+isin θ)^(1/3) , v=(1/( (√)3))(cos θ−isin θ)^(1/3) where tan θ=((√((4/(27))−c^2 ))/c) u=(1/( (√)3)){cos (((θ+2kπ))/3)+isin (((θ+2kπ))/3)},v=(1/( (√3))){cos (((θ+2kπ))/3)−isin (((θ+2kπ))/3)},k=0,1,2 For each value of k we will get an ordered pair(u,v) each of which corresponds to a value of x, θ=tan^(−1) ((√((4/(27))−c^2 ))/c)
$$ \\ $$$$\left({p}+{q}\right)^{\mathrm{3}} −\mathrm{3}{pq}\left({p}+{q}\right)={c} \\ $$$${x}^{\mathrm{3}} −{x}−{c}=\mathrm{0},{x}={p}+{q} \\ $$$${x}={u}+{v} \\ $$$$\Rightarrow{x}^{\mathrm{3}} ={u}^{\mathrm{3}} +{v}^{\mathrm{3}} +\mathrm{3}{uvx} \\ $$$$\Rightarrow{x}^{\mathrm{3}} −\mathrm{3}{uvx}−\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$$\therefore{uv}=\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{and}\:{u}^{\mathrm{3}} +{v}^{\mathrm{3}} ={c} \\ $$$$\left({u}^{\mathrm{3}} −{v}^{\mathrm{3}} \right)^{\mathrm{2}} =\left({u}^{\mathrm{3}} +{v}^{\mathrm{3}} \right)^{\mathrm{2}} −\mathrm{4}{u}^{\mathrm{3}} {v}^{\mathrm{3}} \\ $$$$\Rightarrow\left({u}^{\mathrm{3}} −{v}^{\mathrm{3}} \right)=\pm\sqrt{{c}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{27}}} \\ $$$$\therefore{u}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left({c}\pm\sqrt{{c}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{27}}}\:\right),{v}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left({c}\mp\sqrt{{c}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{27}}}\right) \\ $$$$\because{u}^{\mathrm{3}} {v}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{27}},\:{u}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left({c}+\sqrt{{c}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{27}}}\:\right),{v}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left({c}−\sqrt{{c}^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{27}}}\:\right) \\ $$$$\because\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}},\:{c}^{\mathrm{2}} <\frac{\mathrm{4}}{\mathrm{27}} \\ $$$$\therefore{u}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left({c}+{i}\sqrt{\frac{\mathrm{4}}{\mathrm{27}}−{c}^{\mathrm{2}} \:}\:\right),{v}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left({c}−{i}\sqrt{\frac{\mathrm{4}}{\mathrm{27}}−{c}^{\mathrm{2}} \:}\:\right) \\ $$$${u}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\mathrm{cos}\:\theta+{i}\mathrm{sin}\:\theta\right)^{\frac{\mathrm{1}}{\mathrm{3}}} ,\:{v}=\frac{\mathrm{1}}{\:\sqrt{}\mathrm{3}}\left(\mathrm{cos}\:\theta−{i}\mathrm{sin}\:\theta\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \mathrm{where}\:\mathrm{tan}\:\theta=\frac{\sqrt{\frac{\mathrm{4}}{\mathrm{27}}−{c}^{\mathrm{2}} }}{{c}} \\ $$$${u}=\frac{\mathrm{1}}{\:\sqrt{}\mathrm{3}}\left\{\mathrm{cos}\:\frac{\left(\theta+\mathrm{2}{k}\pi\right)}{\mathrm{3}}+{i}\mathrm{sin}\:\frac{\left(\theta+\mathrm{2}{k}\pi\right)}{\mathrm{3}}\right\},{v}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left\{\mathrm{cos}\:\frac{\left(\theta+\mathrm{2}{k}\pi\right)}{\mathrm{3}}−{i}\mathrm{sin}\:\frac{\left(\theta+\mathrm{2}{k}\pi\right)}{\mathrm{3}}\right\},{k}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$$\mathrm{For}\:\mathrm{each}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{we}\:\mathrm{will}\:\mathrm{get}\:\mathrm{an}\:\mathrm{ordered}\:\mathrm{pair}\left({u},{v}\right) \\ $$$$\mathrm{each}\:\mathrm{of}\:\mathrm{which}\:\mathrm{corresponds}\:\mathrm{to}\:\mathrm{a}\:\mathrm{value}\:\mathrm{of}\:{x},\: \\ $$$$\theta=\mathrm{tan}^{−\mathrm{1}} \frac{\sqrt{\frac{\mathrm{4}}{\mathrm{27}}−{c}^{\mathrm{2}} }}{{c}} \\ $$$$ \\ $$
Commented by anurup last updated on 17/Feb/23
waiting for your feedback
$$\mathrm{waiting}\:\mathrm{for}\:\mathrm{your}\:\mathrm{feedback} \\ $$
Commented by ajfour last updated on 17/Feb/23
thanks, see i have taken another way...
$${thanks},\:{see}\:{i}\:{have}\:{taken}\:{another} \\ $$$${way}… \\ $$
Answered by ajfour last updated on 17/Feb/23
Basically x^3 =x+c If x=p+q p^3 +q^3 +3pq(p+q)=p+q+c If we take p^3 +q^3 =c ⇒ pq=(1/3) as p+q=x≠0 let another way x=((√3)/2)(((cos φ)/(cos θ))) ⇒ 3(√3)cos^3 φ=((√3)cos φ)(2cos θ)^2 +8ccos^3 θ from 4cos^3 φ−3cos φ=cos 3φ we compare coefficients ((3(√3))/(4(√3)cos^2 θ))=(4/3) ⇒ cos^2 θ=(9/(16)) first simply let cos θ=(3/4) ⇒ ((3(√3))/4)cos 3φ=8ccos^3 θ cos 3φ=(4/(3(√3)))(8c)((3/4))^3 =((3(√3)c)/2) ⇒ 3φ=2kπ±cos^(−1) (((3(√3)c)/2)) ⇒ x=p+q=((√3)/2)×(4/3)cos φ x=(2/( (√3)))cos [((2kπ)/3)±(1/3)cos^(−1) (((3(√3)c)/2))] k=0,1,2 say one p+q=x is then for k=0 x=p+q=(2/( (√3)))cos ((1/3)cos^(−1) (((3(√3)c)/2)))
$${Basically}\:\:{x}^{\mathrm{3}} ={x}+{c} \\ $$$${If}\:{x}={p}+{q} \\ $$$${p}^{\mathrm{3}} +{q}^{\mathrm{3}} +\mathrm{3}{pq}\left({p}+{q}\right)={p}+{q}+{c} \\ $$$${If}\:\:{we}\:{take}\:\:\:{p}^{\mathrm{3}} +{q}^{\mathrm{3}} ={c}\:\:\Rightarrow\:\: \\ $$$${pq}=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:{as}\:\:{p}+{q}={x}\neq\mathrm{0} \\ $$$${let}\:{another}\:{way}\:\:{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\phi}{\mathrm{cos}\:\theta}\right) \\ $$$$\Rightarrow\:\:\mathrm{3}\sqrt{\mathrm{3}}\mathrm{cos}\:^{\mathrm{3}} \phi=\left(\sqrt{\mathrm{3}}\mathrm{cos}\:\phi\right)\left(\mathrm{2cos}\:\theta\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{8}{c}\mathrm{cos}\:^{\mathrm{3}} \theta \\ $$$${from}\:\:\mathrm{4cos}\:^{\mathrm{3}} \phi−\mathrm{3cos}\:\phi=\mathrm{cos}\:\mathrm{3}\phi \\ $$$${we}\:{compare}\:{coefficients} \\ $$$$\:\:\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}\sqrt{\mathrm{3}}\mathrm{cos}\:^{\mathrm{2}} \theta}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\Rightarrow\:\:\mathrm{cos}\:^{\mathrm{2}} \theta=\frac{\mathrm{9}}{\mathrm{16}} \\ $$$${first}\:{simply}\:{let}\:\:\mathrm{cos}\:\theta=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{cos}\:\mathrm{3}\phi=\mathrm{8}{c}\mathrm{cos}\:^{\mathrm{3}} \theta \\ $$$$\mathrm{cos}\:\mathrm{3}\phi=\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}\left(\mathrm{8}{c}\right)\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{3}} =\frac{\mathrm{3}\sqrt{\mathrm{3}}{c}}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{3}\phi=\mathrm{2}{k}\pi\pm\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}\sqrt{\mathrm{3}}{c}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:{x}={p}+{q}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{4}}{\mathrm{3}}\mathrm{cos}\:\phi \\ $$$${x}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{cos}\:\left[\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\pm\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}\sqrt{\mathrm{3}}{c}}{\mathrm{2}}\right)\right] \\ $$$${k}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$${say}\:{one}\:{p}+{q}={x}\:{is}\:{then}\:{for}\:{k}=\mathrm{0} \\ $$$$\:\:\:{x}={p}+{q}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mathrm{cos}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{3}\sqrt{\mathrm{3}}{c}}{\mathrm{2}}\right)\right) \\ $$
Commented by anurup last updated on 17/Feb/23
wow! Innovative, but I just want to know what makes you think to assume x=((√3)/2)(((cos φ)/(cos θ)))
$$\mathrm{wow}!\:\mathrm{Innovative},\:\mathrm{but}\:\mathrm{I}\:\mathrm{just}\:\mathrm{want}\:\mathrm{to}\:\mathrm{know}\:\mathrm{what} \\ $$$$\mathrm{makes}\:\mathrm{you}\:\mathrm{think}\:\mathrm{to}\:\mathrm{assume}\:{x}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\phi}{\mathrm{cos}\:\theta}\right) \\ $$
Commented by ajfour last updated on 17/Feb/23
i have tried 3500 substitutions!
$${i}\:{have}\:{tried}\:\mathrm{3500}\:{substitutions}! \\ $$

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