Question Number 192883 by ajfour last updated on 30/May/23
Answered by ajfour last updated on 30/May/23
![(p+s)^3 −(p+s)=k p^3 −p=c s{(p+s)^2 +p^2 +p(p+s)}−s=k−c 3p^2 +3sp+s^2 =((s+k−c)/s) 3sp^2 =(((s+k−c−s^3 −3s)p)/s)−3c =s+k−c−s^3 −3s^2 p p=((s(k+s+2c−s^3 ))/(2s^3 −2s−c+k)) 3s^2 (−s^3 +s+2c+k)^2 + 3s^2 (−s^3 +s+2c+k)(2s^3 −2s−c+k) =(−s^3 +s−c+k)(2s^3 −2s−c+k)^2 if s^3 =(c/2) 3s^2 (s+((3c)/2)+k)^2 +3s^2 (s+((3c)/2)+k)(k−2s) =(s−((3c)/2)+k)(k−2s)^2 p=((s(k+s+2c−s^3 ))/(2s^3 −2s−c+k)) =((s(s+((3c)/2)+k))/(k+s−3s)) ⇒ p(k+s−3s)=s(k+s+((3c)/2)) k+s=((3sp+((3cs)/2))/(p−s))=((3s(p+(c/2)))/(p−s)) 3p^2 +3sp=s−((3c)/2)+k =((3s(p+(c/2))−((3cp)/2)+((3cs)/2))/(p−s)) 3p(p+s) =((3p(s−(c/2))+3cs)/(p−s)) p(p^2 −s^2 )=p(s−(c/2))+cs p^3 −{s(s+1)−(c/2)}p−cs=0 ((p/(2s)))^3 −{(1/4)[1+((2/c))^(1/3) ]−(1/8)}(p/(2s))−(1/4)((c/2))^(1/3) =0 ....](https://www.tinkutara.com/question/Q192899.png)
$$\left({p}+{s}\right)^{\mathrm{3}} −\left({p}+{s}\right)={k} \\ $$$${p}^{\mathrm{3}} −{p}={c} \\ $$$${s}\left\{\left({p}+{s}\right)^{\mathrm{2}} +{p}^{\mathrm{2}} +{p}\left({p}+{s}\right)\right\}−{s}={k}−{c} \\ $$$$\mathrm{3}{p}^{\mathrm{2}} +\mathrm{3}{sp}+{s}^{\mathrm{2}} =\frac{{s}+{k}−{c}}{{s}} \\ $$$$\mathrm{3}{sp}^{\mathrm{2}} =\frac{\left({s}+{k}−{c}−{s}^{\mathrm{3}} −\mathrm{3}{s}\right){p}}{{s}}−\mathrm{3}{c} \\ $$$$\:\:\:\:\:\:={s}+{k}−{c}−{s}^{\mathrm{3}} −\mathrm{3}{s}^{\mathrm{2}} {p} \\ $$$${p}=\frac{{s}\left({k}+{s}+\mathrm{2}{c}−{s}^{\mathrm{3}} \right)}{\mathrm{2}{s}^{\mathrm{3}} −\mathrm{2}{s}−{c}+{k}} \\ $$$$\mathrm{3}{s}^{\mathrm{2}} \left(−{s}^{\mathrm{3}} +{s}+\mathrm{2}{c}+{k}\right)^{\mathrm{2}} + \\ $$$$\mathrm{3}{s}^{\mathrm{2}} \left(−{s}^{\mathrm{3}} +{s}+\mathrm{2}{c}+{k}\right)\left(\mathrm{2}{s}^{\mathrm{3}} −\mathrm{2}{s}−{c}+{k}\right) \\ $$$$=\left(−{s}^{\mathrm{3}} +{s}−{c}+{k}\right)\left(\mathrm{2}{s}^{\mathrm{3}} −\mathrm{2}{s}−{c}+{k}\right)^{\mathrm{2}} \\ $$$${if}\:\:\:{s}^{\mathrm{3}} =\frac{{c}}{\mathrm{2}}\:\:\: \\ $$$$\mathrm{3}{s}^{\mathrm{2}} \left({s}+\frac{\mathrm{3}{c}}{\mathrm{2}}+{k}\right)^{\mathrm{2}} +\mathrm{3}{s}^{\mathrm{2}} \left({s}+\frac{\mathrm{3}{c}}{\mathrm{2}}+{k}\right)\left({k}−\mathrm{2}{s}\right) \\ $$$$\:\:\:\:\:=\left({s}−\frac{\mathrm{3}{c}}{\mathrm{2}}+{k}\right)\left({k}−\mathrm{2}{s}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${p}=\frac{{s}\left({k}+{s}+\mathrm{2}{c}−{s}^{\mathrm{3}} \right)}{\mathrm{2}{s}^{\mathrm{3}} −\mathrm{2}{s}−{c}+{k}} \\ $$$$\:\:\:=\frac{{s}\left({s}+\frac{\mathrm{3}{c}}{\mathrm{2}}+{k}\right)}{{k}+{s}−\mathrm{3}{s}} \\ $$$$\Rightarrow\:\:\:{p}\left({k}+{s}−\mathrm{3}{s}\right)={s}\left({k}+{s}+\frac{\mathrm{3}{c}}{\mathrm{2}}\right) \\ $$$${k}+{s}=\frac{\mathrm{3}{sp}+\frac{\mathrm{3}{cs}}{\mathrm{2}}}{{p}−{s}}=\frac{\mathrm{3}{s}\left({p}+\frac{{c}}{\mathrm{2}}\right)}{{p}−{s}} \\ $$$$\mathrm{3}{p}^{\mathrm{2}} +\mathrm{3}{sp}={s}−\frac{\mathrm{3}{c}}{\mathrm{2}}+{k} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{3}{s}\left({p}+\frac{{c}}{\mathrm{2}}\right)−\frac{\mathrm{3}{cp}}{\mathrm{2}}+\frac{\mathrm{3}{cs}}{\mathrm{2}}}{{p}−{s}} \\ $$$$\:\:\:\:\:\:\mathrm{3}{p}\left({p}+{s}\right)\:=\frac{\mathrm{3}{p}\left({s}−\frac{{c}}{\mathrm{2}}\right)+\mathrm{3}{cs}}{{p}−{s}} \\ $$$${p}\left({p}^{\mathrm{2}} −{s}^{\mathrm{2}} \right)={p}\left({s}−\frac{{c}}{\mathrm{2}}\right)+{cs} \\ $$$${p}^{\mathrm{3}} −\left\{{s}\left({s}+\mathrm{1}\right)−\frac{{c}}{\mathrm{2}}\right\}{p}−{cs}=\mathrm{0} \\ $$$$\left(\frac{{p}}{\mathrm{2}{s}}\right)^{\mathrm{3}} −\left\{\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{1}+\left(\frac{\mathrm{2}}{{c}}\right)^{\mathrm{1}/\mathrm{3}} \right]−\frac{\mathrm{1}}{\mathrm{8}}\right\}\frac{{p}}{\mathrm{2}{s}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{0} \\ $$$$…. \\ $$