Question Number 16794 by tawa tawa last updated on 26/Jun/17
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 26/Jun/17
![c−x=t^3 ,x−a=s^3 ⇒x=((s^3 −t^3 )/2)+((a+c)/2) a+c=c−2+c=2(c−1)=2b,c−a=2 ((c−x))^(1/3) −((a−x))^(1/3) =(c−x−x+a)((((c−x)^2 ))^(1/3) +(((c−x)(x−a)))^(1/3) +(((a−x)^2 ))^(1/3) ) ⇒(((t−s)(t^2 −ts+s^2 ))/(c+a))=b−x ⇒t^3 −t^2 s+ts^2 −st^2 +ts^2 −s^3 =(a+c)(b−x) t^3 −s^3 −2st^2 +2ts^2 =2(b)(((t^3 −s^3 )/2)) t^3 −s^3 −2st^2 +2ts^2 =b(t^3 −s^3 ) ⇒(1−b)(t^3 −s^3 )−2ts(t−s)=0 ⇒(t−s)[(1−b)(t^2 +ts+s^2 )−2ts)=0 ⇒1)t−s=0⇒t=s⇒c−x=x−a⇒x=((c+a)/2)=b 2)t=s⇒b−x=0⇒x=b 3) (1−b)t^2 −(1+b)s.t+(1−b)s^2 =0 t=(((1+b)s±(√((1+b)^2 s^2 −4(1−b)^2 s^2 )))/(2(1−b)))= =(((1+b)s±s(√((1+b−2+2b)(1+b+2−2b))))/(2(1−b)))= =(((1+b)s±s(√((3b−1)(3−b))))/(2(1−b)))=(m±n).s [n=((√((3b−1)(3−b)))/(2(1−b))),m=((1+b)/(2(1−b)))] ⇒(t/s)=m±n⇒(t^3 /s^3 )=(m±n)^3 ⇒((c−x)/(x−a))=(m±n)^3 ⇒((2x−(a+c))/(c−a))=((1−(m±n)^3 )/(1+(m±n)^3 )) ⇒x=b+((1−(m±n)^3 )/(1+(m±n)^3 )) x=b+((1−(((1+b±(√((3b−1)(3−b))))/(2(1−b))))^3 )/(1+(((1+b±(√((3b−1)(3−b))))/(2(1−b))))^3 )) .](Q16831.png)
$${c}−{x}={t}^{\mathrm{3}} ,{x}−{a}={s}^{\mathrm{3}} \Rightarrow{x}=\frac{{s}^{\mathrm{3}} −{t}^{\mathrm{3}} }{\mathrm{2}}+\frac{{a}+{c}}{\mathrm{2}} \\ $$$${a}+{c}={c}−\mathrm{2}+{c}=\mathrm{2}\left({c}−\mathrm{1}\right)=\mathrm{2}{b},{c}−{a}=\mathrm{2} \\ $$$$\sqrt[{\mathrm{3}}]{{c}−{x}}−\sqrt[{\mathrm{3}}]{{a}−{x}}=\left({c}−{x}−{x}+{a}\right)\left(\sqrt[{\mathrm{3}}]{\left({c}−{x}\right)^{\mathrm{2}} }+\sqrt[{\mathrm{3}}]{\left({c}−{x}\right)\left({x}−{a}\right)}+\sqrt[{\mathrm{3}}]{\left({a}−{x}\right)^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\frac{\left({t}−{s}\right)\left({t}^{\mathrm{2}} −{ts}+{s}^{\mathrm{2}} \right)}{{c}+{a}}={b}−{x} \\ $$$$\Rightarrow{t}^{\mathrm{3}} −{t}^{\mathrm{2}} {s}+{ts}^{\mathrm{2}} −{st}^{\mathrm{2}} +{ts}^{\mathrm{2}} −{s}^{\mathrm{3}} =\left({a}+{c}\right)\left({b}−{x}\right) \\ $$$${t}^{\mathrm{3}} −{s}^{\mathrm{3}} −\mathrm{2}{st}^{\mathrm{2}} +\mathrm{2}{ts}^{\mathrm{2}} =\mathrm{2}\left({b}\right)\left(\frac{{t}^{\mathrm{3}} −{s}^{\mathrm{3}} }{\mathrm{2}}\right) \\ $$$${t}^{\mathrm{3}} −{s}^{\mathrm{3}} −\mathrm{2}{st}^{\mathrm{2}} +\mathrm{2}{ts}^{\mathrm{2}} ={b}\left({t}^{\mathrm{3}} −{s}^{\mathrm{3}} \right) \\ $$$$\Rightarrow\left(\mathrm{1}−{b}\right)\left({t}^{\mathrm{3}} −{s}^{\mathrm{3}} \right)−\mathrm{2}{ts}\left({t}−{s}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({t}−{s}\right)\left[\left(\mathrm{1}−{b}\right)\left({t}^{\mathrm{2}} +{ts}+{s}^{\mathrm{2}} \right)−\mathrm{2}{ts}\right)=\mathrm{0} \\ $$$$\left.\Rightarrow\mathrm{1}\right){t}−{s}=\mathrm{0}\Rightarrow{t}={s}\Rightarrow{c}−{x}={x}−{a}\Rightarrow{x}=\frac{{c}+{a}}{\mathrm{2}}={b} \\ $$$$\left.\mathrm{2}\right){t}={s}\Rightarrow{b}−{x}=\mathrm{0}\Rightarrow{x}={b} \\ $$$$\left.\mathrm{3}\right)\:\left(\mathrm{1}−{b}\right){t}^{\mathrm{2}} −\left(\mathrm{1}+{b}\right){s}.{t}+\left(\mathrm{1}−{b}\right){s}^{\mathrm{2}} =\mathrm{0} \\ $$$${t}=\frac{\left(\mathrm{1}+{b}\right){s}\pm\sqrt{\left(\mathrm{1}+{b}\right)^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}−{b}\right)^{\mathrm{2}} {s}^{\mathrm{2}} }}{\mathrm{2}\left(\mathrm{1}−{b}\right)}= \\ $$$$=\frac{\left(\mathrm{1}+{b}\right){s}\pm{s}\sqrt{\left(\mathrm{1}+{b}−\mathrm{2}+\mathrm{2}{b}\right)\left(\mathrm{1}+{b}+\mathrm{2}−\mathrm{2}{b}\right)}}{\mathrm{2}\left(\mathrm{1}−{b}\right)}= \\ $$$$=\frac{\left(\mathrm{1}+{b}\right){s}\pm{s}\sqrt{\left(\mathrm{3}{b}−\mathrm{1}\right)\left(\mathrm{3}−{b}\right)}}{\mathrm{2}\left(\mathrm{1}−{b}\right)}=\left({m}\pm{n}\right).{s} \\ $$$$\left[{n}=\frac{\sqrt{\left(\mathrm{3}{b}−\mathrm{1}\right)\left(\mathrm{3}−{b}\right)}}{\mathrm{2}\left(\mathrm{1}−{b}\right)},{m}=\frac{\mathrm{1}+{b}}{\mathrm{2}\left(\mathrm{1}−{b}\right)}\right] \\ $$$$\Rightarrow\frac{{t}}{{s}}={m}\pm{n}\Rightarrow\frac{{t}^{\mathrm{3}} }{{s}^{\mathrm{3}} }=\left({m}\pm{n}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\frac{{c}−{x}}{{x}−{a}}=\left({m}\pm{n}\right)^{\mathrm{3}} \Rightarrow\frac{\mathrm{2}{x}−\left({a}+{c}\right)}{{c}−{a}}=\frac{\mathrm{1}−\left({m}\pm{n}\right)^{\mathrm{3}} }{\mathrm{1}+\left({m}\pm{n}\right)^{\mathrm{3}} } \\ $$$$\Rightarrow{x}={b}+\frac{\mathrm{1}−\left({m}\pm{n}\right)^{\mathrm{3}} }{\mathrm{1}+\left({m}\pm{n}\right)^{\mathrm{3}} } \\ $$$${x}={b}+\frac{\mathrm{1}−\left(\frac{\mathrm{1}+{b}\pm\sqrt{\left(\mathrm{3}{b}−\mathrm{1}\right)\left(\mathrm{3}−{b}\right)}}{\mathrm{2}\left(\mathrm{1}−{b}\right)}\right)^{\mathrm{3}} }{\mathrm{1}+\left(\frac{\mathrm{1}+{b}\pm\sqrt{\left(\mathrm{3}{b}−\mathrm{1}\right)\left(\mathrm{3}−{b}\right)}}{\mathrm{2}\left(\mathrm{1}−{b}\right)}\right)^{\mathrm{3}} }\:. \\ $$
Commented by tawa tawa last updated on 26/Jun/17
$$\mathrm{wow}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$