Question Number 141794 by mathdanisur last updated on 23/May/21
Commented by JDamian last updated on 23/May/21
$${repeated}\:{question}\:\mathrm{141730} \\ $$
Commented by mathdanisur last updated on 23/May/21
$${solution},\:{no}\:{answer} \\ $$
Answered by MJS_new last updated on 23/May/21
$$\mathrm{my}\:\mathrm{path}: \\ $$$$\mathrm{let}\:{y}={px}\wedge{z}={qx} \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right){x}^{\mathrm{2}} =\mathrm{9} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\left({p}^{\mathrm{2}} +{pq}+{q}^{\mathrm{2}} \right){x}^{\mathrm{2}} =\mathrm{16} \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:\left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right){x}^{\mathrm{2}} =\mathrm{25} \\ $$$$\Rightarrow \\ $$$$\frac{{p}^{\mathrm{2}} +{p}+\mathrm{1}}{\mathrm{9}}=\frac{{p}^{\mathrm{2}} +{pq}+{q}^{\mathrm{2}} }{\mathrm{16}}=\frac{{q}^{\mathrm{2}} +{q}+\mathrm{1}}{\mathrm{25}} \\ $$$$\left({a}\right)=\left({b}\right)=\left({c}\right) \\ $$$$\mathrm{from}\:\mathrm{solving}\:\left({a}\right)=\left({b}\right)\:\wedge\:\left({b}\right)=\left({c}\right)\:\mathrm{we}\:\mathrm{get} \\ $$$${q}=\frac{−{p}\left(\mathrm{2}{p}+\mathrm{1}\right)}{{p}−\mathrm{1}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\left({a}\right)=\left({c}\right) \\ $$$$\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)\left({p}^{\mathrm{2}} +\frac{\mathrm{32}}{\mathrm{11}}{p}−\frac{\mathrm{16}}{\mathrm{11}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{p}=−\frac{\mathrm{16}+\mathrm{12}\sqrt{\mathrm{3}}}{\mathrm{11}}\vee{p}=\frac{−\mathrm{16}+\mathrm{12}\sqrt{\mathrm{3}}}{\mathrm{11}} \\ $$$$\mathrm{now}\:\mathrm{insert}\:\mathrm{backwards} \\ $$
Answered by mr W last updated on 23/May/21
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{9}\:\:\:\:\:…\left({i}\right) \\ $$$${y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{yz}=\mathrm{16}\:\:\:\:…\left({ii}\right) \\ $$$${z}^{\mathrm{2}} +{x}^{\mathrm{2}} +{zx}=\mathrm{25}\:\:\:\:…\left({iii}\right) \\ $$$$\left({ii}\right)−\left({i}\right): \\ $$$${z}^{\mathrm{2}} −{x}^{\mathrm{2}} +{y}\left({z}−{x}\right)=\mathrm{7} \\ $$$$\left({z}−{x}\right)\left({x}+{y}+{z}\right)=\mathrm{7} \\ $$$${let}\:{s}={x}+{y}+{z} \\ $$$$\Rightarrow{z}=\frac{\mathrm{7}}{{s}}+{x} \\ $$$$\left({iii}\right)−\left({ii}\right): \\ $$$$\left({x}−{y}\right)\left({x}+{y}+{z}\right)=\mathrm{9} \\ $$$$\Rightarrow{y}=−\frac{\mathrm{9}}{{s}}+{x} \\ $$$${x}+{y}+{z}={x}−\frac{\mathrm{9}}{{s}}+{x}+\frac{\mathrm{7}}{{s}}+{x}=\mathrm{3}{x}−\frac{\mathrm{2}}{{s}}={s} \\ $$$$\Rightarrow{x}=\frac{{s}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}{s}} \\ $$$$\Rightarrow{y}=−\frac{\mathrm{9}}{{s}}+\frac{{s}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}{s}}=\frac{{s}}{\mathrm{3}}−\frac{\mathrm{25}}{\mathrm{3}{s}} \\ $$$${insert}\:{into}\:\left({i}\right): \\ $$$$\left(\frac{{s}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}{s}}\right)^{\mathrm{2}} +\left(\frac{{s}}{\mathrm{3}}−\frac{\mathrm{25}}{\mathrm{3}{s}}\right)^{\mathrm{2}} +\left(\frac{{s}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}{s}}\right)\left(\frac{{s}}{\mathrm{3}}−\frac{\mathrm{25}}{\mathrm{3}{s}}\right)=\mathrm{9} \\ $$$${s}^{\mathrm{2}} −\mathrm{50}+\frac{\mathrm{193}}{{s}^{\mathrm{2}} }=\mathrm{0} \\ $$$${s}^{\mathrm{4}} −\mathrm{50}{s}^{\mathrm{2}} +\mathrm{193}=\mathrm{0} \\ $$$$\Rightarrow{s}^{\mathrm{2}} =\mathrm{25}\pm\mathrm{12}\sqrt{\mathrm{3}} \\ $$$$ \\ $$$$\left({i}\right)+\left({ii}\right)+\left({iii}\right): \\ $$$$\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+\left({zy}+{yz}+{zx}\right)=\mathrm{50} \\ $$$$ \\ $$$${s}^{\mathrm{2}} =\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +\mathrm{2}\left({xy}+{yz}+{zx}\right) \\ $$$$\mathrm{2}{s}^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+{xy}+{yz}+{zx}+\mathrm{3}\left({xy}+{yz}+{zx}\right) \\ $$$$\mathrm{2}\left(\mathrm{25}\pm\mathrm{12}\sqrt{\mathrm{3}}\right)=\mathrm{50}+\mathrm{3}\left({xy}+{yz}+{zx}\right) \\ $$$$\Rightarrow{xy}+{yz}+{zx}=\pm\mathrm{8}\sqrt{\mathrm{3}} \\ $$