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Question Number 83721 by mr W last updated on 05/Mar/20

Commented by MJS last updated on 05/Mar/20

$$-3$$

Commented by john santu last updated on 05/Mar/20

$$\left(1\right)+\left(2\right)+\left(3\right)$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}-(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}})=\mathrm{a}+\mathrm{b}+\mathrm{c}$$$$\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}}=0$$$$\frac{\mathrm{a}+\mathrm{b}}{\mathrm{ab}}+\frac{1}{\mathrm{c}}=0$$$$\frac{\mathrm{ac}+\mathrm{bc}+\mathrm{ab}}{\mathrm{abc}}=0\Rightarrow \frac{{\mathrm{c}}^2-1+{\mathrm{b}}^2-1+{\mathrm{a}}^2-1}{\mathrm{abc}}=0$$$${\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2=3$$$${(\mathrm{a}+\mathrm{b}+\mathrm{c})}^2-2(\mathrm{ab}+\mathrm{ac}+\mathrm{bc})=3$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}=\sqrt{3}$$$$$$

Commented by mr W last updated on 05/Mar/20

$$\mathit{S}\mathit{O}\mathit{R}\mathit{R}\mathit{Y}!$$$$-3isthecorrectanswer!$$

Commented by john santu last updated on 05/Mar/20

$$\mathrm{haha}...,{\mathrm{i}}^{\prime }\mathrm{m}\mathrm{focus}\mathrm{on}\mathrm{finding}\mathrm{the}$$$$\mathrm{abc}-\mathrm{value}$$

Commented by MJS last updated on 05/Mar/20

$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow c=-\frac{ab}{a+b}$$$$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\Rightarrow c=-(a+b)$$$$-\frac{ab}{a+b}=-(a+b)\Rightarrow b=a(-\frac{1}{2}\pm \frac{\sqrt{3}}{2}\mathrm{i})$$$$\Rightarrow c=a(-\frac{1}{2}\mp \frac{\sqrt{3}}{2}\mathrm{i})$$$$b-\frac{1}{b}=c$$$$\Rightarrow a^2=-\frac{1}{2}\pm \frac{\sqrt{3}}{6}\mathrm{i}$$$$\mathrm{but}\mathrm{then}a-\frac{1}{a}\ne b\wedge c-\frac{1}{c}\ne a$$

Commented by Prithwish Sen 1 last updated on 05/Mar/20

$$\mathrm{But}\mathrm{Sir},$$$$\mathrm{abc}(\frac{1}{\mathrm{a}}+\frac{1}{\mathrm{b}}+\frac{1}{\mathrm{c}})=\mathrm{ab}+\mathrm{bc}+\mathrm{ac}$$

Commented by MJS last updated on 05/Mar/20

$$\mathrm{see}\mathrm{my}\mathrm{solution}\mathrm{below},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}?$$

Commented by Prithwish Sen 1 last updated on 05/Mar/20

$$a-b=\frac{1}{a}......\left(i\right),b-c=\frac{1}{b}....\left(ii\right),c-a=\frac{1}{c}...\left(iii\right)$$$$\left(\mathbf{i}\right)\times \mathbf{c}+\left(\mathbf{ii}\right)\times \mathbf{a}+\left(\mathbf{iii}\right)\times \mathbf{b}\mathbf{we}\mathbf{get}$$$$\frac{\mathbf{c}}{\mathbf{a}}+\frac{\mathbf{b}}{\mathbf{c}}+\frac{\mathbf{a}}{\mathbf{b}}=0....\left(\mathbf{iv}\right)$$$$\mathbf{again},\left(\mathbf{i}\right)\times \frac{1}{\mathbf{b}}+\left(\mathbf{ii}\right)\times \frac{1}{\mathbf{c}}+\left(\mathbf{iii}\right)\times \frac{1}{\mathbf{a}}\mathbf{we}\mathbf{get}$$$$\frac{1}{\mathbf{ab}}+\frac{1}{\mathbf{bc}}+\frac{1}{\mathbf{ac}}=-3+(\frac{\mathbf{a}}{\mathbf{b}}+\frac{\mathbf{b}}{\mathbf{c}}+\frac{\mathbf{c}}{\mathbf{a}})=-3\left(\mathbf{from}\mathbf{iv}\right)$$

Answered by MJS last updated on 05/Mar/20

$$b=\frac{a^2-1}{a}$$$$\Rightarrow c=\frac{a^4-3a^2+1}{a^3-a}$$$$\Rightarrow a=\frac{a^8-7a^6+13a^4-7a^2+1}{a^7+4a^5+4a^3-a}$$$$\frac{3a^6-9a^4+6a^2-1}{a^7-4a^5+4a^3-a}=0$$$$a^6-3a^4+2a^2-\frac{1}{3}=0$$$$y=a^2$$$$y^3-3y^2+2y-\frac{1}{3}=0$$$$z=y-1$$$$z^3-z-\frac{1}{3}=0$$$$z_1=\frac{2\sqrt{3}}{3}\cos \frac{\pi }{18}$$$$z_2=-\frac{2\sqrt{3}}{3}\sin \frac{2\pi }{9}$$$$z_3=-\frac{2\sqrt{3}}{3}\sin \frac{\pi }{9}$$$$y_1=1+\frac{2\sqrt{3}}{3}\cos \frac{\pi }{18}$$$$y_2=1-\frac{2\sqrt{3}}{3}\sin \frac{2\pi }{9}$$$$y_3=1-\frac{2\sqrt{3}}{3}\sin \frac{\pi }{9}$$$$a_{1,4}=\pm \sqrt{1+\frac{2\sqrt{3}}{3}\cos \frac{\pi }{18}}$$$$a_{2,5}=\pm \sqrt{1-\frac{2\sqrt{3}}{3}\sin \frac{2\pi }{9}}$$$$a_{3,6}=\pm \sqrt{1-\frac{2\sqrt{3}}{3}\sin \frac{\pi }{9}}$$$$\mathrm{with}\mathrm{these}\mathrm{values}$$$$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=-3$$

Commented by Prithwish Sen 1 last updated on 05/Mar/20

$$\mathrm{perfect}\mathrm{sir}.\mathrm{How}\mathrm{r}\mathrm{u}\mathrm{sir}.$$

Answered by mr W last updated on 05/Mar/20

Answered by behi83417@gmail.com last updated on 05/Mar/20

$$\frac{\mathrm{a}}{\mathrm{b}}-\frac{1}{\mathrm{ab}}=1,\frac{\mathrm{b}}{\mathrm{c}}-\frac{1}{\mathrm{bc}}=1,\frac{\mathrm{c}}{\mathrm{a}}-\frac{1}{\mathrm{ac}}=1$$$$\Rightarrow \frac{1}{\mathrm{ab}}+\frac{1}{\mathrm{bc}}+\frac{1}{\mathrm{ca}}=(\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}})-3=-3$$$$[\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}}=\mathrm{c}(\mathrm{a}-\mathrm{b})+\mathrm{a}(\mathrm{b}-\mathrm{c})+\mathrm{b}(\mathrm{c}-\mathrm{a})=0]$$

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