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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

$$c-x=t^3,x-a=s^3\Rightarrow x=\frac{s^3-t^3}{2}+\frac{a+c}{2}$$$$a+c=c-2+c=2(c-1)=2b,c-a=2$$$$\sqrt[3]{c-x}-\sqrt[3]{a-x}=(c-x-x+a)(\sqrt[3]{{(c-x)}^2}+\sqrt[3]{(c-x)(x-a)}+\sqrt[3]{{(a-x)}^2})$$$$\Rightarrow \frac{(t-s)(t^2-ts+s^2)}{c+a}=b-x$$$$\Rightarrow t^3-t^{2}s+{ts}^2-{st}^2+{ts}^2-s^3=(a+c)(b-x)$$$$t^3-s^3-2{st}^2+2{ts}^2=2\left(b\right)\left(\frac{t^3-s^3}{2}\right)$$$$t^3-s^3-2{st}^2+2{ts}^2=b(t^3-s^3)$$$$\Rightarrow (1-b)(t^3-s^3)-2ts(t-s)=0$$$$\Rightarrow (t-s)[(1-b)(t^2+ts+s^2)-2ts)=0$$$$\Rightarrow 1)t-s=0\Rightarrow t=s\Rightarrow c-x=x-a\Rightarrow x=\frac{c+a}{2}=b$$$$2)t=s\Rightarrow b-x=0\Rightarrow x=b$$$$3)(1-b)t^2-(1+b)s.t+(1-b)s^2=0$$$$t=\frac{(1+b)s\pm \sqrt{{(1+b)}^2{s}^2-4{(1-b)}^2{s}^2}}{2(1-b)}=$$$$=\frac{(1+b)s\pm s\sqrt{(1+b-2+2b)(1+b+2-2b)}}{2(1-b)}=$$$$=\frac{(1+b)s\pm s\sqrt{(3b-1)(3-b)}}{2(1-b)}=(m\pm n).s$$$$[n=\frac{\sqrt{(3b-1)(3-b)}}{2(1-b)},m=\frac{1+b}{2(1-b)}]$$$$\Rightarrow \frac{t}{s}=m\pm n\Rightarrow \frac{t^3}{s^3}={(m\pm n)}^3$$$$\Rightarrow \frac{c-x}{x-a}={(m\pm n)}^3\Rightarrow \frac{2x-(a+c)}{c-a}=\frac{1-{(m\pm n)}^3}{1+{(m\pm n)}^3}$$$$\Rightarrow x=b+\frac{1-{(m\pm n)}^3}{1+{(m\pm n)}^3}$$$$x=b+\frac{1-{\left(\frac{1+b\pm \sqrt{(3b-1)(3-b)}}{2(1-b)}\right)}^3}{1+{\left(\frac{1+b\pm \sqrt{(3b-1)(3-b)}}{2(1-b)}\right)}^3}.$$

Commented by tawa tawa last updated on 26/Jun/17

$$\mathrm{wow}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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