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Question Number 48559 by behi83417@gmail.com last updated on 25/Nov/18

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18

$$z(x^2-yz)=a^{2}z$$$$x(y^2-xz)=b^{2}x$$$$y(z^2-xy)=c^{2}y$$$$addthem$$$$a^{2}z+b^{2}x+c^{2}y=0$$$$y(x^2-yz)=a^{2}y$$$$z(y^2-xz)=b^{2}z$$$$x(z^2-xy)=c^{2}x$$$$addthem$$$$a^{2}y+b^{2}z+c^{2}x=0$$$$a^{2}z+b^{2}x+c^{2}y=0$$$$c^{2}x+a^{2}y+b^{2}z=0$$$$b^{2}x+c^{2}y+a^{2}z=0$$$$\frac{x}{a^4-b^2{c}^2}=\frac{y}{b^4-a^2{c}^2}=\frac{z}{c^4-a^2{b}^2}=k\left(say\right)$$$$\frac{x}{p}=\frac{y}{q}=\frac{z}{r}=k$$$$puttingthevalueofxyandzin$$$$x^2-yz=a^2$$$$p^2{k}^2-qk\times rk=a^2$$$$k^2(p^2-qr)=a^2$$$$k^2=\frac{a^2}{p^2-qr}k=\pm \frac{a}{\sqrt{p^2-qr}}$$$$x=pksox=\pm \frac{ap}{\sqrt{p^2-qr}}$$$$y=qksoy=\pm \frac{aq}{\sqrt{p^2-qr}}$$$$z=rksoz=\pm \frac{ar}{\sqrt{p^2-qr}}$$$$nowcalculatingvalueof{p}^2-qr$$$${(a^4-b^2{c}^2)}^2-(b^4-a^2{c}^2)(c^4-a^2{b}^2)$$$$=(a^8-2a^4{b}^2{c}^2+b^4{c}^4)-(b^4{c}^4-a^2{b}^6-a^2{c}^6+a^4{b}^2{c}^2)$$$$=a^8-3a^4{b}^2{c}^2+a^2{b}^6+a^2{c}^6$$$$=a^2(a^6+b^6+c^6-3a^2{b}^2{c}^2)$$$$so\sqrt{p^2-qr}=a{(a^6+b^6+c^6-3a^2{b}^2{c}^2)}^{\frac{1}{2}}$$$$x=\pm \frac{ap}{\sqrt{p^2-qr}}=\pm \frac{a(a^4-b^2{c}^2)}{a\sqrt{a^6+b^6+c^6-3a^2{b}^2{c}^2}}$$$$x=\pm \frac{(a^4-b^2{c}^2)}{\sqrt{a^6+b^6+c^6-3a^2{b}^2{c}^2}}$$$$y=\pm \frac{(b^4-a^2{c}^2)}{\sqrt{a^6+b^6+c^6-3a^2{b}^2{c}^2}}$$$$z=\pm \frac{(c^4-a^2{b}^2)}{\sqrt{a^6+b^6+c^6-3a^2{b}^2{c}^2}}$$$$plscheck...$$$$$$$$$$$$$$$$$$

Commented by behi83417@gmail.com last updated on 25/Nov/18

$$rightanswersir.thankyouverymuch.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18

$$thankyousir...$$

Answered by ajfour last updated on 25/Nov/18

$$(x-y)(x+y+z)=a^2-b^2$$$$(y-z)(x+y+z)=b^2-c^2$$$$(z-x)(x+y+z)=c^2-a^2$$$$$$$$\Rightarrow \frac{x-y}{y-z}=\frac{a^2-b^2}{b^2-c^2}=p$$$$\frac{y-z}{z-x}=\frac{b^2-c^2}{c^2-a^2}=q$$$$\frac{z-x}{x-y}=\frac{c^2-a^2}{a^2-b^2}=r$$$$Nowlet\frac{x}{y}=u,\frac{z}{y}=v$$$$whereu\ne 1\mathrm{\&}v\ne 1$$$$\Rightarrow u-1=p(1-v)...\left(i\right)$$$$\mathrm{\&}1-v=q(v-u)...\left(ii\right)$$$$\mathrm{\&}v-u=r(u-1)...\left(iii\right)$$$$rewriting\left(i\right)\mathrm{\&}\left(ii\right)$$$$u+pv=p+1$$$$-qu+(q+1)v=1$$$$\Rightarrow \mathit{v}=\frac{\mathit{q}(\mathit{p}+1)+1}{\mathit{p}\mathit{q}+\mathit{q}+1}=1$$$$\Rightarrow \mathit{u}=1$$$$\Rightarrow u,vnotdefined$$$$\Rightarrow \mathit{y}=0$$$$So{\mathit{x}}^2={\mathit{a}}^2,{\mathit{z}}^2={\mathit{c}}^2,\mathrm{\&}\mathit{x}\mathit{z}=-{\mathit{b}}^2$$$$similarlyif\mathit{x}=0,then$$$${\mathit{y}}^2={\mathit{b}}^2,{\mathit{z}}^2={\mathit{c}}^2,\mathit{y}\mathit{z}=-{\mathit{a}}^2$$$$Andif\mathit{z}=0$$$${\mathit{x}}^2={\mathit{a}}^2,{\mathit{y}}^2={\mathit{b}}^2,\mathit{x}\mathit{y}=-{\mathit{c}}^2.$$

Answered by behi83417@gmail.com last updated on 25/Nov/18

$$let:r=x+y+z$$$$r.(x-y)=a^2-b^2\left(i\right)$$$$r.(y-z)=b^2-c^2\left(ii\right)$$$$r.(z-x)=c^2-a^2\left(iii\right)$$$$1.if:a^2=b^2=c^2\Rightarrow \{\begin{array}{l}[x=y=z\Rightarrow a=b=c=0]\\ [r=x+y+z=0\Rightarrow x^2+xy+y^2=a^2]\ast \end{array}$$$$from(\ast )wecantakeanyvalveforx$$$$andthenobtainothervaribles.$$$$2.if:a^2,or{b}^2,or{c}^2,or{a}^2+b^2+c^2\ne 0,wehave:$$$$r.y=r.x-(a^2-b^2)$$$$r.z=r.x+(c^2-a^2)$$$$\Rightarrow r^2=r.(x+y+z)=3r.x+(b^2+c^2-2a^2)$$$$or:3r.x=r^2+(2a^2-b^2-c^2)\left(\mathbf{X}\right)$$$$3r.y=r^2+(2b^2-a^2-c^2)\left(\mathbf{Y}\right)$$$$3r.z=r^2+(2c^2-a^2-b^2)\left(\mathbf{Z}\right)$$$$alsowehave:$$$${(x-y)}^2+{(y-x)}^2+{(z-x)}^2=2(a^2+b^2+c^2)$$$$nowfrom(i,ii,iii)andthis\uparrow have:$$$${\mathit{r}}^2=\frac{{({\mathit{a}}^2-{\mathit{b}}^2)}^2+{({\mathit{b}}^2-{\mathit{c}}^2)}^2+{({\mathit{c}}^2-{\mathit{a}}^2)}^2}{2({\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2)}\stackrel{aftersymplifing}{=}$$$$=\frac{{\mathit{a}}^6+{\mathit{b}}^6+{\mathit{c}}^6-3{\mathit{a}}^2{\mathit{b}}^2{\mathit{c}}^2}{{({\mathit{a}}^2+{\mathit{b}}^2+{\mathit{c}}^2)}^2}$$$$nowfrom:\left(\mathbf{X}\right)\Rightarrow$$$$x=\frac{1}{3r}[r^2+(2a^2-b^2-c^2)]\stackrel{aftersymplifing}{=}$$$$=\frac{{\mathit{a}}^4-{\mathit{b}}^2{\mathit{c}}^2}{\sqrt{{\mathit{a}}^6+{\mathit{b}}^6+{\mathit{c}}^6-3{\mathit{a}}^2{\mathit{b}}^2{\mathit{c}}^2}}.$$$$\mathit{a}\mathit{n}\mathit{d}\mathit{f}\mathit{i}\mathit{n}\mathit{a}\mathit{l}\mathit{l}\mathit{y}:$$$$\mathbf{x}=\frac{{\mathbf{a}}^4-{\mathbf{b}}^2{\mathbf{c}}^2}{\sqrt{{\mathbf{a}}^6+{\mathbf{b}}^6+{\mathbf{c}}^6-3{\mathbf{a}}^2{\mathbf{b}}^2{\mathbf{c}}^2}}$$$$\mathbf{y}=\frac{{\mathbf{b}}^4-{\mathbf{a}}^2{\mathbf{c}}^2}{\sqrt{{\mathbf{a}}^6+{\mathbf{b}}^6+{\mathbf{c}}^6-3{\mathbf{a}}^2{\mathbf{b}}^2{\mathbf{c}}^2}}$$$$\mathbf{z}=\frac{{\mathbf{c}}^4-{\mathbf{a}}^2{\mathbf{b}}^2}{\sqrt{{\mathbf{a}}^6+{\mathbf{b}}^6+{\mathbf{c}}^6-3{\mathbf{a}}^2{\mathbf{b}}^2{\mathbf{c}}^2}}.$$

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