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
$${z}\left({x}^{\mathrm{2}} −{yz}\right)={a}^{\mathrm{2}} {z} \\ $$$${x}\left({y}^{\mathrm{2}} −{xz}\right)={b}^{\mathrm{2}} {x} \\ $$$${y}\left({z}^{\mathrm{2}} −{xy}\right)={c}^{\mathrm{2}} {y} \\ $$$${add}\:{them} \\ $$$${a}^{\mathrm{2}} {z}+{b}^{\mathrm{2}} {x}+{c}^{\mathrm{2}} {y}=\mathrm{0} \\ $$$${y}\left({x}^{\mathrm{2}} −{yz}\right)={a}^{\mathrm{2}} {y} \\ $$$${z}\left({y}^{\mathrm{2}} −{xz}\right)={b}^{\mathrm{2}} {z} \\ $$$${x}\left({z}^{\mathrm{2}} −{xy}\right)={c}^{\mathrm{2}} {x} \\ $$$${add}\:{them}\: \\ $$$${a}^{\mathrm{2}} {y}+{b}^{\mathrm{2}} {z}+{c}^{\mathrm{2}} {x}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} {z}+{b}^{\mathrm{2}} {x}+{c}^{\mathrm{2}} {y}=\mathrm{0} \\ $$$${c}^{\mathrm{2}} {x}+{a}^{\mathrm{2}} {y}+{b}^{\mathrm{2}} {z}=\mathrm{0} \\ $$$${b}^{\mathrm{2}} {x}+{c}^{\mathrm{2}} {y}+{a}^{\mathrm{2}} {z}=\mathrm{0} \\ $$$$\frac{{x}}{{a}^{\mathrm{4}} −{b}^{\mathrm{2}} {c}^{\mathrm{2}} }=\frac{{y}}{{b}^{\mathrm{4}} −{a}^{\mathrm{2}} {c}^{\mathrm{2}} }=\frac{{z}}{{c}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} }={k}\left({say}\right) \\ $$$$\frac{{x}}{{p}}=\frac{{y}}{{q}}=\frac{{z}}{{r}}={k} \\ $$$${putting}\:{the}\:{value}\:{of}\:{x}\:{y}\:{and}\:{z}\:{in} \\ $$$${x}^{\mathrm{2}} −{yz}={a}^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} {k}^{\mathrm{2}} −{qk}×{rk}={a}^{\mathrm{2}} \\ $$$${k}^{\mathrm{2}} \left({p}^{\mathrm{2}} −{qr}\right)={a}^{\mathrm{2}} \\ $$$${k}^{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{{p}^{\mathrm{2}} −{qr}}\:\:\:\:{k}=\pm\frac{{a}}{\:\sqrt{{p}^{\mathrm{2}} −{qr}}} \\ $$$${x}={pk}\:\:\:\:{so}\:{x}=\pm\frac{{ap}}{\:\sqrt{{p}^{\mathrm{2}} −{qr}}} \\ $$$${y}={qk}\:\:\:{so}\:{y}=\pm\frac{{aq}}{\:\sqrt{{p}^{\mathrm{2}} −{qr}}} \\ $$$${z}={rk}\:\:\:{so}\:{z}=\pm\frac{{ar}}{\:\sqrt{{p}^{\mathrm{2}} −{qr}}} \\ $$$${now}\:{calculating}\:{value}\:{of}\:{p}^{\mathrm{2}} −{qr} \\ $$$$\left({a}^{\mathrm{4}} −{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({b}^{\mathrm{4}} −{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right)\left({c}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right) \\ $$$$=\left({a}^{\mathrm{8}} −\mathrm{2}{a}^{\mathrm{4}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{4}} {c}^{\mathrm{4}} \right)−\left({b}^{\mathrm{4}} {c}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{6}} −{a}^{\mathrm{2}} {c}^{\mathrm{6}} +{a}^{\mathrm{4}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \right) \\ $$$$={a}^{\mathrm{8}} −\mathrm{3}{a}^{\mathrm{4}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{6}} +{a}^{\mathrm{2}} {c}^{\mathrm{6}} \\ $$$$={a}^{\mathrm{2}} \left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} −\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \right) \\ $$$${so}\:\sqrt{{p}^{\mathrm{2}} −{qr}}\:={a}\left({a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} −\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$${x}=\pm\frac{{ap}}{\:\sqrt{{p}^{\mathrm{2}} −{qr}}\:}=\pm\frac{{a}\left({a}^{\mathrm{4}} −{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)}{{a}\sqrt{{a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} −\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }\:} \\ $$$${x}=\pm\frac{\left({a}^{\mathrm{4}} −{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)}{\:\sqrt{{a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} −\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }\:} \\ $$$${y}=\pm\frac{\left({b}^{\mathrm{4}} −{a}^{\mathrm{2}} {c}^{\mathrm{2}} \right)}{\:\sqrt{{a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} −\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }\:} \\ $$$${z}=\pm\frac{\left({c}^{\mathrm{4}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} \right)}{\:\sqrt{{a}^{\mathrm{6}} +{b}^{\mathrm{6}} +{c}^{\mathrm{6}} −\mathrm{3}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }} \\ $$$${pls}\:{check}… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by behi83417@gmail.com last updated on 25/Nov/18
$${right}\:{answer}\:{sir}.{thank}\:{you}\:{very}\:{much}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18
$${thank}\:{you}\:{sir}… \\ $$
Answered by ajfour last updated on 25/Nov/18
$$\left({x}−{y}\right)\left({x}+{y}+{z}\right)\:=\:{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \: \\ $$$$\left({y}−{z}\right)\left({x}+{y}+{z}\right)\:=\:{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$$\left({z}−{x}\right)\left({x}+{y}+{z}\right)\:=\:{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow\:\:\:\frac{{x}−{y}}{{y}−{z}}\:=\:\frac{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }\:=\:{p} \\ $$$$\:\:\:\:\:\:\:\:\frac{{y}−{z}}{{z}−{x}}\:=\:\frac{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:=\:{q} \\ $$$$\:\:\:\:\:\:\:\:\frac{{z}−{x}}{{x}−{y}}\:=\:\frac{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\:=\:{r} \\ $$$${Now}\:\:{let}\:\:\frac{{x}}{{y}}\:={u}\:\:,\:\frac{{z}}{{y}}\:=\:{v} \\ $$$${where}\:\:{u}\neq\mathrm{1}\:\&\:{v}\neq\:\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:{u}−\mathrm{1}=\:{p}\left(\mathrm{1}−{v}\right)\:\:\:\:\:\:\:\:\:\:…\left({i}\right) \\ $$$$\&\:\:\:\:\:\:\mathrm{1}−{v}\:=\:{q}\left({v}−{u}\right)\:\:\:\:\:\:\:\:…\left({ii}\right) \\ $$$$\&\:\:\:\:\:\:{v}−{u}\:=\:{r}\left({u}−\mathrm{1}\right)\:\:\:\:\:\:\:\:…\left({iii}\right) \\ $$$$\:{rewriting}\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{u}+{pv}\:=\:{p}+\mathrm{1} \\ $$$$\:\:\:\:−{qu}+\left({q}+\mathrm{1}\right){v}\:=\:\mathrm{1} \\ $$$$\:\Rightarrow\:\:\:\boldsymbol{{v}}\:=\:\frac{\boldsymbol{{q}}\left(\boldsymbol{{p}}+\mathrm{1}\right)+\mathrm{1}}{\boldsymbol{{pq}}+\boldsymbol{{q}}+\mathrm{1}}\:=\:\mathrm{1}\:\: \\ $$$$\:\Rightarrow\:\:\:\boldsymbol{{u}}\:=\:\mathrm{1} \\ $$$$\:\Rightarrow\:\:{u},\:{v}\:{not}\:{defined} \\ $$$$\Rightarrow\:\:\boldsymbol{{y}}\:=\:\mathrm{0}\: \\ $$$${So}\:\boldsymbol{{x}}^{\mathrm{2}} \:=\:\boldsymbol{{a}}^{\mathrm{2}} ,\:\:\boldsymbol{{z}}^{\mathrm{2}} \:=\:\boldsymbol{{c}}^{\mathrm{2}} \:,\:\&\:\boldsymbol{{xz}}\:=\:−\boldsymbol{{b}}^{\mathrm{2}} \\ $$$${similarly}\:{if}\:\:\boldsymbol{{x}}=\:\mathrm{0},\:{then} \\ $$$$\:\:\:\:\:\:\boldsymbol{{y}}^{\mathrm{2}} \:=\:\boldsymbol{{b}}^{\mathrm{2}} \:,\:\boldsymbol{{z}}^{\mathrm{2}} \:=\:\boldsymbol{{c}}^{\mathrm{2}} \:,\:\boldsymbol{{yz}}\:=\:−\boldsymbol{{a}}^{\mathrm{2}} \\ $$$${And}\:{if}\:\:\boldsymbol{{z}}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{2}} \:=\:\boldsymbol{{a}}^{\mathrm{2}} ,\:\boldsymbol{{y}}^{\mathrm{2}} \:=\:\boldsymbol{{b}}^{\mathrm{2}} ,\:\boldsymbol{{xy}}\:=\:−\boldsymbol{{c}}^{\mathrm{2}} \:\:. \\ $$
Answered by behi83417@gmail.com last updated on 25/Nov/18
$${let}:\:{r}={x}+{y}+{z} \\ $$$${r}.\left({x}−{y}\right)={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \:\left({i}\right) \\ $$$${r}.\left({y}−{z}\right)={b}^{\mathrm{2}} −{c}^{\mathrm{2}} \:\left({ii}\right) \\ $$$${r}.\left({z}−{x}\right)={c}^{\mathrm{2}} −{a}^{\mathrm{2}} \:\left({iii}\right) \\ $$$$\mathrm{1}.{if}:{a}^{\mathrm{2}} ={b}^{\mathrm{2}} ={c}^{\mathrm{2}} \Rightarrow\begin{cases}{\left[{x}={y}={z}\Rightarrow{a}={b}={c}=\mathrm{0}\right]}\\{\left[{r}={x}+{y}+{z}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} ={a}^{\mathrm{2}} \right]\ast}\end{cases} \\ $$$${from}\left(\ast\right){we}\:{can}\:{take}\:{any}\:{valve}\:{for}\:{x} \\ $$$${and}\:{then}\:{obtain}\:{other}\:{varibles}. \\ $$$$\mathrm{2}.{if}:\:{a}^{\mathrm{2}} ,{or}\:{b}^{\mathrm{2}} ,{or}\:{c}^{\mathrm{2}} ,{or}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \neq\mathrm{0},{we}\:{have}: \\ $$$${r}.{y}={r}.{x}−\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$${r}.{z}={r}.{x}+\left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{r}^{\mathrm{2}} ={r}.\left({x}+{y}+{z}\right)=\mathrm{3}{r}.{x}+\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} \right) \\ $$$${or}:\:\mathrm{3}{r}.{x}={r}^{\mathrm{2}} +\left(\mathrm{2}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\:\left(\boldsymbol{\mathrm{X}}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{3}{r}.{y}={r}^{\mathrm{2}} +\left(\mathrm{2}{b}^{\mathrm{2}} −{a}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\:\left(\boldsymbol{\mathrm{Y}}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{3}{r}.{z}={r}^{\mathrm{2}} +\left(\mathrm{2}{c}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\:\left(\boldsymbol{\mathrm{Z}}\right) \\ $$$${also}\:{we}\:{have}: \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} +\left({y}−{x}\right)^{\mathrm{2}} +\left({z}−{x}\right)^{\mathrm{2}} =\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\: \\ $$$${now}\:{from}\:\left({i},{ii},{iii}\right){and}\:{this}\uparrow{have}: \\ $$$$\boldsymbol{{r}}^{\mathrm{2}} =\frac{\left(\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\boldsymbol{{b}}^{\mathrm{2}} −\boldsymbol{{c}}^{\mathrm{2}} \right)^{\mathrm{2}} +\left(\boldsymbol{{c}}^{\mathrm{2}} −\boldsymbol{{a}}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} \right)}\overset{{after}\:{symplifing}} {=} \\ $$$$=\frac{\boldsymbol{{a}}^{\mathrm{6}} +\boldsymbol{{b}}^{\mathrm{6}} +\boldsymbol{{c}}^{\mathrm{6}} −\mathrm{3}\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}^{\mathrm{2}} }{\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${now}\:{from}:\:\left(\boldsymbol{\mathrm{X}}\right)\Rightarrow \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{3}{r}}\left[{r}^{\mathrm{2}} +\left(\mathrm{2}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)\right]\overset{{after}\:{symplifing}} {=} \\ $$$$=\frac{\boldsymbol{{a}}^{\mathrm{4}} −\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}^{\mathrm{2}} }{\:\sqrt{\boldsymbol{{a}}^{\mathrm{6}} +\boldsymbol{{b}}^{\mathrm{6}} +\boldsymbol{{c}}^{\mathrm{6}} −\mathrm{3}\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}^{\mathrm{2}} }}. \\ $$$$\boldsymbol{{and}}\:\boldsymbol{{finally}}: \\ $$$$\boldsymbol{\mathrm{x}}=\frac{\boldsymbol{\mathrm{a}}^{\mathrm{4}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} \boldsymbol{\mathrm{c}}^{\mathrm{2}} }{\:\sqrt{\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{b}}^{\mathrm{6}} +\boldsymbol{\mathrm{c}}^{\mathrm{6}} −\mathrm{3}\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{b}}^{\mathrm{2}} \boldsymbol{\mathrm{c}}^{\mathrm{2}} }} \\ $$$$\boldsymbol{\mathrm{y}}=\frac{\boldsymbol{\mathrm{b}}^{\mathrm{4}} −\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{c}}^{\mathrm{2}} }{\:\sqrt{\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{b}}^{\mathrm{6}} +\boldsymbol{\mathrm{c}}^{\mathrm{6}} −\mathrm{3}\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{b}}^{\mathrm{2}} \boldsymbol{\mathrm{c}}^{\mathrm{2}} }} \\ $$$$\boldsymbol{\mathrm{z}}=\frac{\boldsymbol{\mathrm{c}}^{\mathrm{4}} −\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{b}}^{\mathrm{2}} }{\:\sqrt{\boldsymbol{\mathrm{a}}^{\mathrm{6}} +\boldsymbol{\mathrm{b}}^{\mathrm{6}} +\boldsymbol{\mathrm{c}}^{\mathrm{6}} −\mathrm{3}\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{b}}^{\mathrm{2}} \boldsymbol{\mathrm{c}}^{\mathrm{2}} }}\:\:\:. \\ $$