x^3 −x−c=0 let x=t+h (t−s){t^3 +3ht^2 +(3h^2 −1)t +(h^3 −h−c)}=0 let t=p+q (p+q−s){p^3 +q^3 +3pq(p+q) +3h(p^2 +q^2 )+6hpq +(3h^2 −1)(p+q)+(h^3 −h−c)} = 0 p^4 +q^4 +pq(p^2 +q^2 )+3pq(p+q)^2 +3h(p^2 +q^2 )(p+q)+6hpq(p+q) +(3h^2 −1)(p+q)^2 +(h^3 −h−c)(p+q) −s(p^3 +q^3 )−3pqs(p+q) −3hs(p^2 +q^2 )−6hspq +(3h^2 −1)s(p+q) −s(h^3 −h−c)=0 let p^4 +q^4 =p^2 +q^2 (p^2 +q^2 )^2 −(p^2 +q^2 )−2p^2 q^2 =0 ⇒ p^2 +q^2 =(1/2)±(√((1/4)+2p^2 q^2 )) say z=(1/2)+(√((1/4)+2m^2 )) (p+q)^2 =z+2m ⇒z(1+m+3m+3h^2 −1−3hs) +(z+2m)^(1/2) {3hz+6mh +h^3 −h−c−3ms −s(z−m)+(3h^2 −1)s} +6m^2 +2(3h^2 −1)m−6hsm −s(h^3 −h−c)=0 let (3h−s)z+m(6h−2s) +h^3 −h−c+3sh^2 −s=0 & 4m+3h(h−s)=0 ⇒ 2(3h−s)z−3h(h−s)(3h−s) +3sh^2 −s+h^3 −h−c=0 let s=−3h ⇒ 12hz−80h^3 −h−c=0 & m=−3h^2 ⇒ 54h^4 −6h^2 (3h^2 −1) −54h^4 +3h(h^3 −h−c)=0 ⇒ 15h^4 −3h^2 +3ch=0 And if h≠0 ⇒ 5h^3 −h+c=0 D=(c^2 /(100))−((1/(15)))^3 ⇒ D≥0 if c≥ (2/(3(√(15)))) m=−3h^2 ; s=−3h; And 12hz−80h^3 −h−c=0 ⇒ z=((20h^2 )/3)+(1/(12))+(c/(12h)) p+q=(√(z+2m)) =(√(((2h^2 )/3)+((c+h)/(12h)))) x=t+h=p+q+h x=h+(√(((2h^2 )/3)+((c+h)/(12h)))) but (c/(12h))>−(((2h^2 )/3)+(1/(12))) ⇒ c<−8h^3 −h ⇒ 8h^3 +h+c<0 5h^3 −h+c=0 ⇒ 13h^3 +2c<0 ......


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Question Number 143582 by ajfour last updated on 16/Jun/21
$x^3 −x−c=0 let x=t+h (t−s){t^3 +3ht^2 +(3h^2 −1)t +(h^3 −h−c)}=0 let t=p+q (p+q−s){p^3 +q^3 +3pq(p+q) +3h(p^2 +q^2 )+6hpq +(3h^2 −1)(p+q)+(h^3 −h−c)} = 0 p^4 +q^4 +pq(p^2 +q^2 )+3pq(p+q)^2 +3h(p^2 +q^2 )(p+q)+6hpq(p+q) +(3h^2 −1)(p+q)^2 +(h^3 −h−c)(p+q) −s(p^3 +q^3 )−3pqs(p+q) −3hs(p^2 +q^2 )−6hspq +(3h^2 −1)s(p+q) −s(h^3 −h−c)=0 let p^4 +q^4 =p^2 +q^2 (p^2 +q^2 )^2 −(p^2 +q^2 )−2p^2 q^2 =0 ⇒ p^2 +q^2 =(1/2)±(√((1/4)+2p^2 q^2 )) say z=(1/2)+(√((1/4)+2m^2 )) (p+q)^2 =z+2m ⇒z(1+m+3m+3h^2 −1−3hs) +(z+2m)^(1/2) {3hz+6mh +h^3 −h−c−3ms −s(z−m)+(3h^2 −1)s} +6m^2 +2(3h^2 −1)m−6hsm −s(h^3 −h−c)=0 let (3h−s)z+m(6h−2s) +h^3 −h−c+3sh^2 −s=0 & 4m+3h(h−s)=0 ⇒ 2(3h−s)z−3h(h−s)(3h−s) +3sh^2 −s+h^3 −h−c=0 let s=−3h ⇒ 12hz−80h^3 −h−c=0 & m=−3h^2 ⇒ 54h^4 −6h^2 (3h^2 −1) −54h^4 +3h(h^3 −h−c)=0 ⇒ 15h^4 −3h^2 +3ch=0 And if h≠0 ⇒ 5h^3 −h+c=0 D=(c^2 /(100))−((1/(15)))^3 ⇒ D≥0 if c≥ (2/(3(√(15)))) m=−3h^2 ; s=−3h; And 12hz−80h^3 −h−c=0 ⇒ z=((20h^2 )/3)+(1/(12))+(c/(12h)) p+q=(√(z+2m)) =(√(((2h^2 )/3)+((c+h)/(12h)))) x=t+h=p+q+h x=h+(√(((2h^2 )/3)+((c+h)/(12h)))) but (c/(12h))>−(((2h^2 )/3)+(1/(12))) ⇒ c<−8h^3 −h ⇒ 8h^3 +h+c<0 5h^3 −h+c=0 ⇒ 13h^3 +2c<0 ......$
$${x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$${let}\:\:{x}={t}+{h} \\ $$$$\left({t}−{s}\right)\left\{{t}^{\mathrm{3}} +\mathrm{3}{ht}^{\mathrm{2}} +\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){t}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({h}^{\mathrm{3}} −{h}−{c}\right)\right\}=\mathrm{0} \\ $$$${let}\:{t}={p}+{q} \\ $$$$\left({p}+{q}−{s}\right)\left\{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} +\mathrm{3}{pq}\left({p}+{q}\right)\right. \\ $$$$\:\:+\mathrm{3}{h}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)+\mathrm{6}{hpq} \\ $$$$\left.\:\:+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)\left({p}+{q}\right)+\left({h}^{\mathrm{3}} −{h}−{c}\right)\right\} \\ $$$$\:\:=\:\mathrm{0} \\ $$$${p}^{\mathrm{4}} +{q}^{\mathrm{4}} +{pq}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)+\mathrm{3}{pq}\left({p}+{q}\right)^{\mathrm{2}} \\ $$$$+\mathrm{3}{h}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left({p}+{q}\right)+\mathrm{6}{hpq}\left({p}+{q}\right) \\ $$$$+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)\left({p}+{q}\right)^{\mathrm{2}} \\ $$$$+\left({h}^{\mathrm{3}} −{h}−{c}\right)\left({p}+{q}\right) \\ $$$$−{s}\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)−\mathrm{3}{pqs}\left({p}+{q}\right) \\ $$$$−\mathrm{3}{hs}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{6}{hspq} \\ $$$$+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){s}\left({p}+{q}\right) \\ $$$$−{s}\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$$${let}\:\:{p}^{\mathrm{4}} +{q}^{\mathrm{4}} ={p}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$$$\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)−\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:{p}^{\mathrm{2}} +{q}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{2}} } \\ $$$${say}\:\:\:\:{z}=\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}{m}^{\mathrm{2}} } \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} ={z}+\mathrm{2}{m} \\ $$$$\Rightarrow{z}\left(\mathrm{1}+{m}+\mathrm{3}{m}+\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}−\mathrm{3}{hs}\right) \\ $$$$+\left({z}+\mathrm{2}{m}\right)^{\mathrm{1}/\mathrm{2}} \left\{\mathrm{3}{hz}+\mathrm{6}{mh}\right. \\ $$$$\:\:\:\:+{h}^{\mathrm{3}} −{h}−{c}−\mathrm{3}{ms} \\ $$$$\left.\:\:\:\:−{s}\left({z}−{m}\right)+\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){s}\right\} \\ $$$$+\mathrm{6}{m}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right){m}−\mathrm{6}{hsm} \\ $$$$−{s}\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$$${let}\:\:\:\left(\mathrm{3}{h}−{s}\right){z}+{m}\left(\mathrm{6}{h}−\mathrm{2}{s}\right) \\ $$$$\:\:\:\:\:\:\:+{h}^{\mathrm{3}} −{h}−{c}+\mathrm{3}{sh}^{\mathrm{2}} −{s}=\mathrm{0} \\ $$$$\&\:\:\mathrm{4}{m}+\mathrm{3}{h}\left({h}−{s}\right)=\mathrm{0}\:\:\Rightarrow \\ $$$$\:\mathrm{2}\left(\mathrm{3}{h}−{s}\right){z}−\mathrm{3}{h}\left({h}−{s}\right)\left(\mathrm{3}{h}−{s}\right) \\ $$$$\:\:+\mathrm{3}{sh}^{\mathrm{2}} −{s}+{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$${let}\:\:{s}=−\mathrm{3}{h} \\ $$$$\Rightarrow\:\mathrm{12}{hz}−\mathrm{80}{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$$\&\:\:{m}=−\mathrm{3}{h}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{54}{h}^{\mathrm{4}} −\mathrm{6}{h}^{\mathrm{2}} \left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:−\mathrm{54}{h}^{\mathrm{4}} +\mathrm{3}{h}\left({h}^{\mathrm{3}} −{h}−{c}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{15}{h}^{\mathrm{4}} −\mathrm{3}{h}^{\mathrm{2}} +\mathrm{3}{ch}=\mathrm{0} \\ $$$${And}\:{if}\:{h}\neq\mathrm{0}\:\:\Rightarrow \\ $$$$\:\:\mathrm{5}{h}^{\mathrm{3}} −{h}+{c}=\mathrm{0} \\ $$$${D}=\frac{{c}^{\mathrm{2}} }{\mathrm{100}}−\left(\frac{\mathrm{1}}{\mathrm{15}}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\:\:{D}\geqslant\mathrm{0}\:\:{if}\:{c}\geqslant\:\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{15}}} \\ $$$${m}=−\mathrm{3}{h}^{\mathrm{2}} \:;\:{s}=−\mathrm{3}{h};\:\:{And} \\ $$$$\:\mathrm{12}{hz}−\mathrm{80}{h}^{\mathrm{3}} −{h}−{c}=\mathrm{0} \\ $$$$\Rightarrow\:\:{z}=\frac{\mathrm{20}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{12}}+\frac{{c}}{\mathrm{12}{h}} \\ $$$${p}+{q}=\sqrt{{z}+\mathrm{2}{m}} \\ $$$$\:\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{2}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{{c}+{h}}{\mathrm{12}{h}}} \\ $$$${x}={t}+{h}={p}+{q}+{h} \\ $$$$\:\:\:{x}={h}+\sqrt{\frac{\mathrm{2}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{{c}+{h}}{\mathrm{12}{h}}} \\ $$$${but}\:\:\:\frac{{c}}{\mathrm{12}{h}}>−\left(\frac{\mathrm{2}{h}^{\mathrm{2}} }{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{12}}\right) \\ $$$$\Rightarrow\:\:{c}<−\mathrm{8}{h}^{\mathrm{3}} −{h} \\ $$$$\Rightarrow\:\:\mathrm{8}{h}^{\mathrm{3}} +{h}+{c}<\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{5}{h}^{\mathrm{3}} −{h}+{c}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{13}{h}^{\mathrm{3}} +\mathrm{2}{c}<\mathrm{0} \\ $$$$...... \\ $$
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