Question Number 50278 by LYCON TRIX last updated on 15/Dec/18
$$\sqrt{\mathrm{a}+\sqrt{\mathrm{a}−\mathrm{x}}}\:+\:\sqrt{\mathrm{a}−\sqrt{\mathrm{a}+\mathrm{x}}}\:=\:\mathrm{2x} \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{beg}\:\mathrm{u}\:\mathrm{guys}\: \\ $$$$\mathrm{please}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{question} \\ $$
Commented by ajfour last updated on 15/Dec/18
$${a}\:{and}\:{x}\:\in\:\mathbb{R}\:\:{or}\:\mathbb{C}\:? \\ $$
Commented by MJS last updated on 15/Dec/18
$$\mathrm{would}\:\mathrm{you}\:\mathrm{be}\:\mathrm{so}\:\mathrm{kind}\:\mathrm{to}\:\mathrm{tell}\:\mathrm{us}\:\mathrm{where}\:\mathrm{this} \\ $$$$\mathrm{equation}\:\mathrm{comes}\:\mathrm{from}? \\ $$
Commented by maxmathsup by imad last updated on 15/Dec/18
$${its}\:{only}\:{a}\:{trying}\:{let}\:{use}\:{the}\:{changement}\:{x}\:={acos}\left(\mathrm{2}{t}\right) \\ $$$$\left({e}\right)\:\Leftrightarrow\:\sqrt{{a}+\sqrt{{a}−{acos}\left(\mathrm{2}{t}\right)}}+\sqrt{{a}−\sqrt{{a}}}=\mathrm{2}{a}\:{cos}\left(\mathrm{2}{t}\right)\:\Rightarrow \\ $$$$\sqrt{{a}+\sqrt{{a}}\sqrt{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}}+\sqrt{{a}−\sqrt{{a}}\sqrt{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}}=\mathrm{2}{a}\:{cos}\left(\mathrm{2}{t}\right)\:\Rightarrow \\ $$$$\sqrt{{a}+\sqrt{\mathrm{2}{a}}{sint}}+\sqrt{{a}−\sqrt{\mathrm{2}{a}}{cost}}=\mathrm{2}{acos}\left(\mathrm{2}{t}\right)\:\:\Rightarrow \\ $$$${a}+\sqrt{\mathrm{2}{a}}{sint}\:+\mathrm{2}\sqrt{{a}+\sqrt{\mathrm{2}{a}}{sint}}\sqrt{{a}−\sqrt{\mathrm{2}{a}}{cost}}\:+{a}−\sqrt{\mathrm{2}{a}}{cost}\:=\mathrm{4}{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\:\Rightarrow \\ $$$$\left.\mathrm{2}{a}+\sqrt{\mathrm{2}{a}}\left({sint}\:−{cost}\right)\:+\mathrm{2}\sqrt{\left({a}+\sqrt{\mathrm{2}{a}}{sint}\right)\left({a}−\sqrt{\mathrm{2}{a}}{cost}\right.}\right)\:=\mathrm{4}{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right)\:\Rightarrow \\ $$$$\left(\mathrm{4}{a}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\mathrm{2}{t}\right)−\mathrm{2}{a}−\sqrt{\mathrm{2}{a}}\left({sint}\:−{cost}\right)\right)^{\mathrm{2}} =\mathrm{4}\left({a}^{\mathrm{2}} −{a}\sqrt{\mathrm{2}{a}}{cost}\:+{a}\sqrt{\mathrm{2}{a}}{sint}\:−\mathrm{2}{asint}\:{cost}\right) \\ $$$$….{be}\:{cntinued}\:\:\:{i}\:{think}\:{that}\:{way}\:{can}\:{give}\:{something}\:{if}\:{we}\:{resist}\:{to}\:{storm} \\ $$$${of}\:{calculus}… \\ $$
Commented by ajfour last updated on 23/Aug/23
$${A}\:{past}\:{prize}\:{question},\:{i}\:{wonder}\:{who}\:{got}\:{the}\:{prize}. \\ $$$${just}\:{name}\:{the}\:{guy},\:{wont}\:{u}? \\ $$
Commented by ajfour last updated on 15/Mar/25
$${okay},\:{i}\:{now}\:{know}\:{how}\:{much}\:{i}\:{need}. \\ $$$$..{say}\:\mathrm{50}\:{crore}\:{rupees}\:\left(\:{Indian}\:{currency}\right) \\ $$$${would}\:{amply}\:{suffice}. \\ $$
Commented by ajfour last updated on 15/Mar/25
Answered by behi83417@gmail.com last updated on 17/Dec/18
![a+x=t^2 ,a−x=s^2 ⇒x=((t^2 −s^2 )/2),a=((t^2 +s^2 )/2) (√(a+s))+(√(a−t))=t^2 −s^2 a+s+a−t+2(√((a+s)(a−t)))=(t^2 −s^2 )^2 2(√((a+s)(a−t)))=(t^2 −s^2 )^2 +t−s−2a= =t^4 +s^4 −2t^2 s^2 −t^2 −s^2 +t−s ⇒4(a^2 −at+as−st)= =t^8 +s^8 +4t^4 s^4 +t^4 +s^4 +t^2 +s^2 + +2t^4 s^4 −4t^6 s^2 −2t^6 −2t^4 s^2 +2t^5 −2t^4 s −4t^2 s^6 −2s^4 t^2 −2s^6 +2ts^4 −2s^5 +4t^4 s^2 +4t^2 s^4 −4t^3 s^2 +4t^2 s^3 +2t^2 s^2 −2t^3 +2t^2 s −2s^2 t+2s^3 −2ts LHS=(t^2 +s^2 )^2 −2(t^2 +s^2 )(t−s)−4st= =t^4 +s^4 +2t^2 s^2 −2t^3 +2t^2 s−2s^2 t+2s^3 −4st RHS=t^8 +s^8 +6t^4 s^4 +t^4 +s^4 +t^2 +s^2 −4t^6 s^2 −2t^6 −2s^6 +2t^4 s^2 +2t^5 −2t^4 s −4t^2 s^6 −2s^4 t^2 +2ts^4 −2s^5 +4t^2 s^4 −4t^3 s^2 +4t^2 s^3 +2t^2 s^2 −2t^3 +2t^2 s−2s^2 t+2s^3 −2ts ⇒t^8 +s^8 −2t^6 −2s^6 +2t^5 +2s^5 +t^2 +s^2 +2ts−4t^6 s^2 +2t^4 s^2 −2t^4 s−4t^2 s^6 +2s^4 t^2 +2ts^4 −4t^3 s^2 +4t^2 s^3 +2t^2 s −2s^2 t=0 (t^8 +s^8 )−2(t^6 +s^6 )+2(t^5 −s^5 )− −4t^2 s^2 (t^4 +s^4 )−2ts(t^3 −s^3 )+ +(2t^2 s^2 +1)(t^2 +s^2 )−4t^2 s^2 (t−s)+ +6t^4 s^4 +2ts=0 t−s=p,ts=q t^8 +s^8 =(t^4 −s^4 )^2 +2t^4 s^4 = =(t^2 −s^2 )^2 (t^2 +s^2 )^2 +2t^4 s^4 = =(t−s)^2 (t+s)^2 (t^2 +s^2 )^2 +2t^4 s^4 = =(t−s)^2 ((t−s)^2 +4ts)((t−s)^2 +2ts)^2 +2t^4 s^4 = =(p)^2 (p^2 +4q)(p^4 +4p^2 q+4q^2 )+2q^4 = =(p^4 +4p^2 q)(p^4 +4p^2 q+4q^2 )+2q^4 = =p^8 +8p^6 q+20p^4 q^2 +16p^2 q^3 +2q^4 t^6 +s^6 =(t^3 −s^3 )^2 +2t^3 s^3 =(t−s)^2 (t^2 +ts+s^2 )^2 +2t^3 s^3 = =(p)^2 ((t−s)^2 +3ts)^2 +2t^3 s^3 = =p^2 (p^2 +3q)^2 +2q^3 = =p^6 +6p^4 q+9p^2 q^2 +2q^3 t^5 −s^5 =(t−s)(t^4 +t^3 s+t^2 s^2 +ts^3 +s^4 )= =(t−s)(t^4 +s^4 +ts(t^2 +ts+s^2 ))= =(t−s)((t^2 +s^2 )^2 −2t^2 s^2 +ts((t−s)^2 +3ts)= =(t−s)[((t−s)^2 +2ts)^2 −2t^2 s^2 +ts((t−s)^2 +3ts))= =p[((p^2 +2q)^2 −2q^2 +q(p^2 +3q))= =p[(p^4 +4p^2 q+4q^2 )−2q^2 +qp^2 +3pq]= =p^5 +5p^2 q+5pq^2 t^4 +s^4 =(t^2 +s^2 )^2 −2t^2 s^2 =[(t−s)^2 +2ts]^2 −2t^2 s^2 = =[p^2 +2q]^2 −2q^2 =p^4 +4p^2 q+2q^2 t^3 −s^3 =(t−s)(t^2 +s^2 +ts)=p^3 +3pq t^2 +s^2 =(t−s)^2 +2ts=p^2 +2q after replacing and simplifing: (p^8 −2p^6 +2p^5 +p^2 )+16p^2 (p^2 −1)q^2 +(8p^6 −12p^4 +8p^3 +4)q=0](https://www.tinkutara.com/question/Q50305.png)
$${a}+{x}={t}^{\mathrm{2}} ,{a}−{x}={s}^{\mathrm{2}} \Rightarrow{x}=\frac{{t}^{\mathrm{2}} −{s}^{\mathrm{2}} }{\mathrm{2}},{a}=\frac{{t}^{\mathrm{2}} +{s}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\sqrt{{a}+{s}}+\sqrt{{a}−{t}}={t}^{\mathrm{2}} −{s}^{\mathrm{2}} \\ $$$${a}+{s}+{a}−{t}+\mathrm{2}\sqrt{\left({a}+{s}\right)\left({a}−{t}\right)}=\left({t}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{2}\sqrt{\left({a}+{s}\right)\left({a}−{t}\right)}=\left({t}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)^{\mathrm{2}} +{t}−{s}−\mathrm{2}{a}= \\ $$$$={t}^{\mathrm{4}} +{s}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} −{t}^{\mathrm{2}} −{s}^{\mathrm{2}} +{t}−{s} \\ $$$$\Rightarrow\mathrm{4}\left({a}^{\mathrm{2}} −{at}+{as}−{st}\right)= \\ $$$$={t}^{\mathrm{8}} +{s}^{\mathrm{8}} +\mathrm{4}{t}^{\mathrm{4}} {s}^{\mathrm{4}} +{t}^{\mathrm{4}} +{s}^{\mathrm{4}} +{t}^{\mathrm{2}} +{s}^{\mathrm{2}} + \\ $$$$+\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{6}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{6}} −\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{5}} −\mathrm{2}{t}^{\mathrm{4}} {s} \\ $$$$−\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{6}} −\mathrm{2}{s}^{\mathrm{4}} {t}^{\mathrm{2}} −\mathrm{2}{s}^{\mathrm{6}} +\mathrm{2}{ts}^{\mathrm{4}} −\mathrm{2}{s}^{\mathrm{5}} +\mathrm{4}{t}^{\mathrm{4}} {s}^{\mathrm{2}} \\ $$$$+\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{3}} {s}^{\mathrm{2}} +\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s} \\ $$$$−\mathrm{2}{s}^{\mathrm{2}} {t}+\mathrm{2}{s}^{\mathrm{3}} −\mathrm{2}{ts} \\ $$$${LHS}=\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)\left({t}−{s}\right)−\mathrm{4}{st}= \\ $$$$={t}^{\mathrm{4}} +{s}^{\mathrm{4}} +\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s}−\mathrm{2}{s}^{\mathrm{2}} {t}+\mathrm{2}{s}^{\mathrm{3}} −\mathrm{4}{st} \\ $$$${RHS}={t}^{\mathrm{8}} +{s}^{\mathrm{8}} +\mathrm{6}{t}^{\mathrm{4}} {s}^{\mathrm{4}} +{t}^{\mathrm{4}} +{s}^{\mathrm{4}} +{t}^{\mathrm{2}} +{s}^{\mathrm{2}} \\ $$$$−\mathrm{4}{t}^{\mathrm{6}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{6}} −\mathrm{2}{s}^{\mathrm{6}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{5}} −\mathrm{2}{t}^{\mathrm{4}} {s} \\ $$$$−\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{6}} −\mathrm{2}{s}^{\mathrm{4}} {t}^{\mathrm{2}} +\mathrm{2}{ts}^{\mathrm{4}} −\mathrm{2}{s}^{\mathrm{5}} +\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{3}} {s}^{\mathrm{2}} \\ $$$$+\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s}−\mathrm{2}{s}^{\mathrm{2}} {t}+\mathrm{2}{s}^{\mathrm{3}} −\mathrm{2}{ts} \\ $$$$\Rightarrow{t}^{\mathrm{8}} +{s}^{\mathrm{8}} −\mathrm{2}{t}^{\mathrm{6}} −\mathrm{2}{s}^{\mathrm{6}} +\mathrm{2}{t}^{\mathrm{5}} +\mathrm{2}{s}^{\mathrm{5}} \\ $$$$+{t}^{\mathrm{2}} +{s}^{\mathrm{2}} +\mathrm{2}{ts}−\mathrm{4}{t}^{\mathrm{6}} {s}^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{4}} {s}−\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{6}} \\ $$$$+\mathrm{2}{s}^{\mathrm{4}} {t}^{\mathrm{2}} +\mathrm{2}{ts}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{3}} {s}^{\mathrm{2}} +\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} {s} \\ $$$$−\mathrm{2}{s}^{\mathrm{2}} {t}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{8}} +{s}^{\mathrm{8}} \right)−\mathrm{2}\left({t}^{\mathrm{6}} +{s}^{\mathrm{6}} \right)+\mathrm{2}\left({t}^{\mathrm{5}} −{s}^{\mathrm{5}} \right)− \\ $$$$−\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{2}} \left({t}^{\mathrm{4}} +{s}^{\mathrm{4}} \right)−\mathrm{2}{ts}\left({t}^{\mathrm{3}} −{s}^{\mathrm{3}} \right)+ \\ $$$$+\left(\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)−\mathrm{4}{t}^{\mathrm{2}} {s}^{\mathrm{2}} \left({t}−{s}\right)+ \\ $$$$+\mathrm{6}{t}^{\mathrm{4}} {s}^{\mathrm{4}} +\mathrm{2}{ts}=\mathrm{0} \\ $$$${t}−{s}={p},{ts}={q} \\ $$$${t}^{\mathrm{8}} +{s}^{\mathrm{8}} =\left({t}^{\mathrm{4}} −{s}^{\mathrm{4}} \right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{4}} = \\ $$$$=\left({t}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{4}} = \\ $$$$=\left({t}−{s}\right)^{\mathrm{2}} \left({t}+{s}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{4}} = \\ $$$$=\left({t}−{s}\right)^{\mathrm{2}} \left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{4}{ts}\right)\left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{2}{ts}\right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{4}} {s}^{\mathrm{4}} = \\ $$$$=\left({p}\right)^{\mathrm{2}} \left({p}^{\mathrm{2}} +\mathrm{4}{q}\right)\left({p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{2}} {q}+\mathrm{4}{q}^{\mathrm{2}} \right)+\mathrm{2}{q}^{\mathrm{4}} = \\ $$$$=\left({p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{2}} {q}\right)\left({p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{2}} {q}+\mathrm{4}{q}^{\mathrm{2}} \right)+\mathrm{2}{q}^{\mathrm{4}} = \\ $$$$={p}^{\mathrm{8}} +\mathrm{8}{p}^{\mathrm{6}} {q}+\mathrm{20}{p}^{\mathrm{4}} {q}^{\mathrm{2}} +\mathrm{16}{p}^{\mathrm{2}} {q}^{\mathrm{3}} +\mathrm{2}{q}^{\mathrm{4}} \\ $$$${t}^{\mathrm{6}} +{s}^{\mathrm{6}} =\left({t}^{\mathrm{3}} −{s}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{3}} {s}^{\mathrm{3}} =\left({t}−{s}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} +{ts}+{s}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{3}} {s}^{\mathrm{3}} = \\ $$$$=\left({p}\right)^{\mathrm{2}} \left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{3}{ts}\right)^{\mathrm{2}} +\mathrm{2}{t}^{\mathrm{3}} {s}^{\mathrm{3}} = \\ $$$$={p}^{\mathrm{2}} \left({p}^{\mathrm{2}} +\mathrm{3}{q}\right)^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{3}} = \\ $$$$={p}^{\mathrm{6}} +\mathrm{6}{p}^{\mathrm{4}} {q}+\mathrm{9}{p}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{2}{q}^{\mathrm{3}} \\ $$$${t}^{\mathrm{5}} −{s}^{\mathrm{5}} =\left({t}−{s}\right)\left({t}^{\mathrm{4}} +{t}^{\mathrm{3}} {s}+{t}^{\mathrm{2}} {s}^{\mathrm{2}} +{ts}^{\mathrm{3}} +{s}^{\mathrm{4}} \right)= \\ $$$$=\left({t}−{s}\right)\left({t}^{\mathrm{4}} +{s}^{\mathrm{4}} +{ts}\left({t}^{\mathrm{2}} +{ts}+{s}^{\mathrm{2}} \right)\right)= \\ $$$$=\left({t}−{s}\right)\left(\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} +{ts}\left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{3}{ts}\right)=\right. \\ $$$$=\left({t}−{s}\right)\left[\left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{2}{ts}\right)^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} +{ts}\left(\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{3}{ts}\right)\right)= \\ $$$$={p}\left[\left(\left({p}^{\mathrm{2}} +\mathrm{2}{q}\right)^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} +{q}\left({p}^{\mathrm{2}} +\mathrm{3}{q}\right)\right)=\right. \\ $$$$={p}\left[\left({p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{2}} {q}+\mathrm{4}{q}^{\mathrm{2}} \right)−\mathrm{2}{q}^{\mathrm{2}} +{qp}^{\mathrm{2}} +\mathrm{3}{pq}\right]= \\ $$$$={p}^{\mathrm{5}} +\mathrm{5}{p}^{\mathrm{2}} {q}+\mathrm{5}{pq}^{\mathrm{2}} \\ $$$${t}^{\mathrm{4}} +{s}^{\mathrm{4}} =\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} =\left[\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{2}{ts}\right]^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{2}} {s}^{\mathrm{2}} = \\ $$$$=\left[{p}^{\mathrm{2}} +\mathrm{2}{q}\right]^{\mathrm{2}} −\mathrm{2}{q}^{\mathrm{2}} ={p}^{\mathrm{4}} +\mathrm{4}{p}^{\mathrm{2}} {q}+\mathrm{2}{q}^{\mathrm{2}} \\ $$$${t}^{\mathrm{3}} −{s}^{\mathrm{3}} =\left({t}−{s}\right)\left({t}^{\mathrm{2}} +{s}^{\mathrm{2}} +{ts}\right)={p}^{\mathrm{3}} +\mathrm{3}{pq} \\ $$$${t}^{\mathrm{2}} +{s}^{\mathrm{2}} =\left({t}−{s}\right)^{\mathrm{2}} +\mathrm{2}{ts}={p}^{\mathrm{2}} +\mathrm{2}{q} \\ $$$${after}\:{replacing}\:{and}\:{simplifing}: \\ $$$$\left({p}^{\mathrm{8}} −\mathrm{2}{p}^{\mathrm{6}} +\mathrm{2}{p}^{\mathrm{5}} +{p}^{\mathrm{2}} \right)+\mathrm{16}{p}^{\mathrm{2}} \left({p}^{\mathrm{2}} −\mathrm{1}\right){q}^{\mathrm{2}} \\ $$$$+\left(\mathrm{8}{p}^{\mathrm{6}} −\mathrm{12}{p}^{\mathrm{4}} +\mathrm{8}{p}^{\mathrm{3}} +\mathrm{4}\right){q}=\mathrm{0} \\ $$
Commented by behi83417@gmail.com last updated on 17/Dec/18
$$\Rightarrow{Q}#\mathrm{50511} \\ $$