Question Number 225392 by Rojarani last updated on 24/Oct/25
Commented by Ghisom_ last updated on 25/Oct/25
$$\mathrm{there}\:\mathrm{should}\:\mathrm{be}\:\mathrm{9}\:\mathrm{solutions}\:\mathrm{for}\:\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:\in\mathbb{C}^{\mathrm{3}} \\ $$
Answered by Kademi last updated on 26/Oct/25
![x^3 +x(y−z)^2 = 2 x(x^2 +x(y−z)^2 ) = 2 x(x^2 +y^2 +z^2 −2yz) = 2 ⇒ x(x^2 +y^2 +z^2 )−2xyz = 2 ⇒ y(x^2 +y^2 +z^2 )−2xyz = 30 ⇒ z(x^2 +y^2 +z^2 )−2xyz = 16 [y(x^2 +y^2 +z^2 )−2xyz]−[z(x^2 +y^2 +z^2 )−2xyz] = 30−16 ⇒ (y−z)(x^2 +y^2 +z^2 ) = 14 ⇒ (z−x)(x^2 +y^2 +z^2 ) = 14 y−z = z−x z = ((x+y)/2) x(x^2 +y^2 +z^2 )−2xyz = 2 x(x^2 +y^2 +(((x+y)/2))^2 )−2xy(((x+y)/2)) = 2 x(((4x^2 +4y^2 +(x^2 +2xy+y^2 ))/4))−x^2 y−xy^2 = 2 (((5x^3 +2x^2 y+5xy^2 )/4))−x^2 y−xy^2 = 2 ((5x^3 +2x^2 y+5xy^2 +(−4x^2 y−4xy^2 ))/4) = 2 ⇒ 5x^3 −2x^2 y+xy^2 = 8 ⇒ 5y^3 −2xy^2 +x^2 y = 120 5x^3 −2x^2 y+xy^2 = 8 75x^3 −30x^2 y+15xy^2 = 120 [75x^3 −30x^2 y+15xy^2 ] = [5y^3 −2xy^2 +x^2 y] 75x^3 −31x^2 y+17xy^2 −5y^3 = 0 75(x^3 /y^3 )−31(x^2 /y^2 )+17(x/y)−5 = 0 t = (x/y) 75t^3 −31t^2 +17t−5 = 0 t = (1/3) ∈ R ; t_(2,3) ∈ C (x/y) = (1/3) y = 3x → z = 2x x^3 +x(y−z)^2 = 2 x^3 +x(3x−2x)^2 = 2 x^3 +x(x)^2 = 2 2x^3 = 2 x^3 = 1 ⇒ x = 1 ⇒ y = 3x → y = 3 ⇒ z = 2x → z = 2 { ((1^3 +1(3−2)^2 = 2)),((3^3 +3(2−1)^2 = 30)),((2^3 +2(1−3)^2 = 16)) :} determinant (((x = 1 ; y = 3 ; z = 2))) Solution by: Harri Savolainen](https://www.tinkutara.com/question/Q225421.png)
$$\:{x}^{\mathrm{3}} +{x}\left({y}−{z}\right)^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\:{x}\left({x}^{\mathrm{2}} +{x}\left({y}−{z}\right)^{\mathrm{2}} \right)\:=\:\mathrm{2} \\ $$$$\:{x}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\mathrm{2}{yz}\right)\:=\:\mathrm{2} \\ $$$$\:\:\:\:\:\Rightarrow\:{x}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\mathrm{2}{xyz}\:=\:\mathrm{2} \\ $$$$\:\:\:\:\:\Rightarrow\:{y}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\mathrm{2}{xyz}\:=\:\mathrm{30} \\ $$$$\:\:\:\:\:\Rightarrow\:{z}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\mathrm{2}{xyz}\:=\:\mathrm{16} \\ $$$$\: \\ $$$$\:\left[{y}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\mathrm{2}{xyz}\right]−\left[{z}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\mathrm{2}{xyz}\right]\:=\:\mathrm{30}−\mathrm{16} \\ $$$$\:\:\:\:\:\Rightarrow\:\left({y}−{z}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\:=\:\mathrm{14} \\ $$$$\:\:\:\:\:\Rightarrow\:\left({z}−{x}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\:=\:\mathrm{14} \\ $$$$\: \\ $$$$\:{y}−{z}\:=\:{z}−{x} \\ $$$$\:{z}\:=\:\frac{{x}+{y}}{\mathrm{2}} \\ $$$$\: \\ $$$$\:{x}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\mathrm{2}{xyz}\:=\:\mathrm{2} \\ $$$$\:{x}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} \right)−\mathrm{2}{xy}\left(\frac{{x}+{y}}{\mathrm{2}}\right)\:=\:\mathrm{2} \\ $$$$\:{x}\left(\frac{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{y}^{\mathrm{2}} +\left({x}^{\mathrm{2}} +\mathrm{2}{xy}+{y}^{\mathrm{2}} \right)}{\mathrm{4}}\right)−{x}^{\mathrm{2}} {y}−{xy}^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\:\left(\frac{\mathrm{5}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} {y}+\mathrm{5}{xy}^{\mathrm{2}} }{\mathrm{4}}\right)−{x}^{\mathrm{2}} {y}−{xy}^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\:\frac{\mathrm{5}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} {y}+\mathrm{5}{xy}^{\mathrm{2}} +\left(−\mathrm{4}{x}^{\mathrm{2}} {y}−\mathrm{4}{xy}^{\mathrm{2}} \right)}{\mathrm{4}}\:=\:\mathrm{2} \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{5}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} {y}+{xy}^{\mathrm{2}} \:=\:\mathrm{8} \\ $$$$\:\:\:\:\:\Rightarrow\:\mathrm{5}{y}^{\mathrm{3}} −\mathrm{2}{xy}^{\mathrm{2}} +{x}^{\mathrm{2}} {y}\:=\:\mathrm{120} \\ $$$$\: \\ $$$$\:\mathrm{5}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} {y}+{xy}^{\mathrm{2}} \:=\:\mathrm{8} \\ $$$$\:\mathrm{75}{x}^{\mathrm{3}} −\mathrm{30}{x}^{\mathrm{2}} {y}+\mathrm{15}{xy}^{\mathrm{2}} \:=\:\mathrm{120} \\ $$$$\:\left[\mathrm{75}{x}^{\mathrm{3}} −\mathrm{30}{x}^{\mathrm{2}} {y}+\mathrm{15}{xy}^{\mathrm{2}} \right]\:=\:\left[\mathrm{5}{y}^{\mathrm{3}} −\mathrm{2}{xy}^{\mathrm{2}} +{x}^{\mathrm{2}} {y}\right] \\ $$$$\:\mathrm{75}{x}^{\mathrm{3}} −\mathrm{31}{x}^{\mathrm{2}} {y}+\mathrm{17}{xy}^{\mathrm{2}} −\mathrm{5}{y}^{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\:\mathrm{75}\frac{{x}^{\mathrm{3}} }{{y}^{\mathrm{3}} }−\mathrm{31}\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }+\mathrm{17}\frac{{x}}{{y}}−\mathrm{5}\:=\:\mathrm{0} \\ $$$$\:{t}\:=\:\frac{{x}}{{y}} \\ $$$$\:\mathrm{75}{t}^{\mathrm{3}} −\mathrm{31}{t}^{\mathrm{2}} +\mathrm{17}{t}−\mathrm{5}\:=\:\mathrm{0} \\ $$$$\:{t}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\in\:\mathbb{R}\:\:;\:\:{t}_{\mathrm{2},\mathrm{3}} \:\in\:\mathbb{C} \\ $$$$\:\frac{{x}}{{y}}\:=\:\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\:{y}\:=\:\mathrm{3}{x}\:\:\rightarrow\:\:{z}\:=\:\mathrm{2}{x} \\ $$$$\: \\ $$$$\:{x}^{\mathrm{3}} +{x}\left({y}−{z}\right)^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\:{x}^{\mathrm{3}} +{x}\left(\mathrm{3}{x}−\mathrm{2}{x}\right)^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\:{x}^{\mathrm{3}} +{x}\left({x}\right)^{\mathrm{2}} \:=\:\mathrm{2} \\ $$$$\:\mathrm{2}{x}^{\mathrm{3}} \:=\:\mathrm{2} \\ $$$$\:{x}^{\mathrm{3}} \:=\:\mathrm{1} \\ $$$$\: \\ $$$$\:\Rightarrow\:\:{x}\:=\:\mathrm{1} \\ $$$$\:\Rightarrow\:{y}\:=\:\mathrm{3}{x}\:\rightarrow\:\:{y}\:=\:\mathrm{3} \\ $$$$\:\Rightarrow\:{z}\:=\:\mathrm{2}{x}\:\:\rightarrow\:\:{z}\:=\:\mathrm{2} \\ $$$$\: \\ $$$$\:\begin{cases}{\mathrm{1}^{\mathrm{3}} +\mathrm{1}\left(\mathrm{3}−\mathrm{2}\right)^{\mathrm{2}} \:=\:\mathrm{2}}\\{\mathrm{3}^{\mathrm{3}} +\mathrm{3}\left(\mathrm{2}−\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{30}}\\{\mathrm{2}^{\mathrm{3}} +\mathrm{2}\left(\mathrm{1}−\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{16}}\end{cases} \\ $$$$\:\:\begin{array}{|c|}{{x}\:=\:\mathrm{1}\:;\:\:{y}\:=\:\mathrm{3}\:\:;\:\:{z}\:=\:\mathrm{2}}\\\hline\end{array} \\ $$$$\: \\ $$$$\:\mathrm{Solution}\:\mathrm{by}:\:\mathrm{Harri}\:\mathrm{Savolainen} \\ $$
Answered by Ghisom_ last updated on 25/Oct/25
![y=px∧z=pqx [obviously x, y, z, p, q ≠0] { (((p^2 (q−1)^2 +1)x^3 =2)),((p(p^2 q^2 +p^2 −2pq+1)x^3 =30)),((pq(p^2 q^2 +(p−1)^2 )x^3 =16)) :} x^3 =(2/(p^2 (q−1)^2 +1))=((30)/(p(p^2 q^2 +p^2 −2pq+1)))=((16)/(pq(p^2 q^2 +(p−1)^2 ))) ⇔ { ((p^3 q^2 +p^3 −15p^2 q^2 +28p^2 q−15p^2 +p−15=0 (I))),((p^3 q^3 +p^3 q−8p^2 q^2 +14p^2 q−8p^2 +pq−8=0 (II))) :} II−qI p^2 (q−1)(15q^2 −21q+8)+15q−8=0 p=±(√(−((15q−8)/((q−1)(15q^2 −21q+8))))) inserting in I & transforming ±(√(−((15q−8)/((q−1)(15q^2 −21q+8)))))=(1/(2q−1)) p=(1/(2q−1)) inserting again to avoid squaring leads to q^3 −((128)/(75))q^2 +((76)/(75))q−((16)/(75))=0 ⇒ q=(2/3)∨q=((13)/(25))±((√(31))/(25))i ⇒ p=3∨p=(1/5)∓((2(√(31)))/5)i ⇒ x^3 =1∧y=3x∧z=2x x^3 =−(1/5)+((2(√(31)))/5)i∧y=((1/5)+((2(√(31)))/5)i)x∧z=((3/5)+((√(31))/5)i)x x^3 =−(1/5)−((2(√(31)))/5)i∧y=((1/5)−((2(√(31)))/5)i)x∧z=((3/5)−((√(31))/5)i)x](https://www.tinkutara.com/question/Q225417.png)
$${y}={px}\wedge{z}={pqx}\:\:\:\:\:\left[\mathrm{obviously}\:{x},\:{y},\:{z},\:{p},\:{q}\:\:\neq\mathrm{0}\right] \\ $$$$\begin{cases}{\left({p}^{\mathrm{2}} \left({q}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{3}} =\mathrm{2}}\\{{p}\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{p}^{\mathrm{2}} −\mathrm{2}{pq}+\mathrm{1}\right){x}^{\mathrm{3}} =\mathrm{30}}\\{{pq}\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +\left({p}−\mathrm{1}\right)^{\mathrm{2}} \right){x}^{\mathrm{3}} =\mathrm{16}}\end{cases} \\ $$$${x}^{\mathrm{3}} =\frac{\mathrm{2}}{{p}^{\mathrm{2}} \left({q}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{30}}{{p}\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +{p}^{\mathrm{2}} −\mathrm{2}{pq}+\mathrm{1}\right)}=\frac{\mathrm{16}}{{pq}\left({p}^{\mathrm{2}} {q}^{\mathrm{2}} +\left({p}−\mathrm{1}\right)^{\mathrm{2}} \right)} \\ $$$$\Leftrightarrow \\ $$$$\begin{cases}{{p}^{\mathrm{3}} {q}^{\mathrm{2}} +{p}^{\mathrm{3}} −\mathrm{15}{p}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{28}{p}^{\mathrm{2}} {q}−\mathrm{15}{p}^{\mathrm{2}} +{p}−\mathrm{15}=\mathrm{0}\:\:\:\:\:\left({I}\right)}\\{{p}^{\mathrm{3}} {q}^{\mathrm{3}} +{p}^{\mathrm{3}} {q}−\mathrm{8}{p}^{\mathrm{2}} {q}^{\mathrm{2}} +\mathrm{14}{p}^{\mathrm{2}} {q}−\mathrm{8}{p}^{\mathrm{2}} +{pq}−\mathrm{8}=\mathrm{0}\:\:\:\:\:\:\:\:\left({II}\right)}\end{cases} \\ $$$${II}−{qI} \\ $$$${p}^{\mathrm{2}} \left({q}−\mathrm{1}\right)\left(\mathrm{15}{q}^{\mathrm{2}} −\mathrm{21}{q}+\mathrm{8}\right)+\mathrm{15}{q}−\mathrm{8}=\mathrm{0} \\ $$$${p}=\pm\sqrt{−\frac{\mathrm{15}{q}−\mathrm{8}}{\left({q}−\mathrm{1}\right)\left(\mathrm{15}{q}^{\mathrm{2}} −\mathrm{21}{q}+\mathrm{8}\right)}} \\ $$$$\mathrm{inserting}\:\mathrm{in}\:{I}\:\&\:\mathrm{transforming} \\ $$$$\pm\sqrt{−\frac{\mathrm{15}{q}−\mathrm{8}}{\left({q}−\mathrm{1}\right)\left(\mathrm{15}{q}^{\mathrm{2}} −\mathrm{21}{q}+\mathrm{8}\right)}}=\frac{\mathrm{1}}{\mathrm{2}{q}−\mathrm{1}} \\ $$$${p}=\frac{\mathrm{1}}{\mathrm{2}{q}−\mathrm{1}} \\ $$$$\mathrm{inserting}\:\mathrm{again}\:\mathrm{to}\:\mathrm{avoid}\:\mathrm{squaring}\:\mathrm{leads}\:\mathrm{to} \\ $$$${q}^{\mathrm{3}} −\frac{\mathrm{128}}{\mathrm{75}}{q}^{\mathrm{2}} +\frac{\mathrm{76}}{\mathrm{75}}{q}−\frac{\mathrm{16}}{\mathrm{75}}=\mathrm{0} \\ $$$$\Rightarrow\:{q}=\frac{\mathrm{2}}{\mathrm{3}}\vee{q}=\frac{\mathrm{13}}{\mathrm{25}}\pm\frac{\sqrt{\mathrm{31}}}{\mathrm{25}}\mathrm{i} \\ $$$$\Rightarrow\:{p}=\mathrm{3}\vee{p}=\frac{\mathrm{1}}{\mathrm{5}}\mp\frac{\mathrm{2}\sqrt{\mathrm{31}}}{\mathrm{5}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{3}} =\mathrm{1}\wedge{y}=\mathrm{3}{x}\wedge{z}=\mathrm{2}{x} \\ $$$${x}^{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{2}\sqrt{\mathrm{31}}}{\mathrm{5}}\mathrm{i}\wedge{y}=\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{2}\sqrt{\mathrm{31}}}{\mathrm{5}}\mathrm{i}\right){x}\wedge{z}=\left(\frac{\mathrm{3}}{\mathrm{5}}+\frac{\sqrt{\mathrm{31}}}{\mathrm{5}}\mathrm{i}\right){x} \\ $$$${x}^{\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{2}\sqrt{\mathrm{31}}}{\mathrm{5}}\mathrm{i}\wedge{y}=\left(\frac{\mathrm{1}}{\mathrm{5}}−\frac{\mathrm{2}\sqrt{\mathrm{31}}}{\mathrm{5}}\mathrm{i}\right){x}\wedge{z}=\left(\frac{\mathrm{3}}{\mathrm{5}}−\frac{\sqrt{\mathrm{31}}}{\mathrm{5}}\mathrm{i}\right){x} \\ $$