Question Number 192409 by Abdullahrussell last updated on 17/May/23
Answered by Frix last updated on 17/May/23
![(x+y+z)^2 =x^3 +y^3 +z^2 −3xyz (x+y+z)^2 =(x+y+z)(x^2 +y^2 +z^2 −(xy+xz+yz)) [1] ★ x+y+z=0 ⇔ x=−(y+z)∧(y, z)∈R^2 [2] x+y+z=x^2 +y^2 +z^2 −(xy+xz+yz) y=px∧z=qx (p+q+1)x=(p^2 −pq+q^2 −p−q+1)x^2 [2.1] ★ x=0 ⇒ y=0∧z=0 [already included above] [2.2] p+q+1=(p^2 −pq+q^2 −p−q+1)x ★ x=((p+q+1)/(p^2 −pq+q^2 −p−q+1))∧y=px∧z=qx∧(p, q)∈R^2 [p^2 −pq+q^2 −p−q+1≠0∀(p, q)∈R^2 ]](https://www.tinkutara.com/question/Q192416.png)
$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{3}} +{y}^{\mathrm{3}} +{z}^{\mathrm{2}} −\mathrm{3}{xyz} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\left({x}+{y}+{z}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\left({xy}+{xz}+{yz}\right)\right) \\ $$$$\left[\mathrm{1}\right] \\ $$$$\bigstar\:{x}+{y}+{z}=\mathrm{0}\:\Leftrightarrow\:{x}=−\left({y}+{z}\right)\wedge\left({y},\:{z}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$$$\left[\mathrm{2}\right] \\ $$$${x}+{y}+{z}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} −\left({xy}+{xz}+{yz}\right) \\ $$$${y}={px}\wedge{z}={qx} \\ $$$$\left({p}+{q}+\mathrm{1}\right){x}=\left({p}^{\mathrm{2}} −{pq}+{q}^{\mathrm{2}} −{p}−{q}+\mathrm{1}\right){x}^{\mathrm{2}} \\ $$$$\left[\mathrm{2}.\mathrm{1}\right] \\ $$$$\bigstar\:{x}=\mathrm{0}\:\Rightarrow\:{y}=\mathrm{0}\wedge{z}=\mathrm{0}\:\left[\mathrm{already}\:\mathrm{included}\:\mathrm{above}\right] \\ $$$$\left[\mathrm{2}.\mathrm{2}\right] \\ $$$${p}+{q}+\mathrm{1}=\left({p}^{\mathrm{2}} −{pq}+{q}^{\mathrm{2}} −{p}−{q}+\mathrm{1}\right){x} \\ $$$$\bigstar\:{x}=\frac{{p}+{q}+\mathrm{1}}{{p}^{\mathrm{2}} −{pq}+{q}^{\mathrm{2}} −{p}−{q}+\mathrm{1}}\wedge{y}={px}\wedge{z}={qx}\wedge\left({p},\:{q}\right)\in\mathbb{R}^{\mathrm{2}} \\ $$$$\left[{p}^{\mathrm{2}} −{pq}+{q}^{\mathrm{2}} −{p}−{q}+\mathrm{1}\neq\mathrm{0}\forall\left({p},\:{q}\right)\in\mathbb{R}^{\mathrm{2}} \right] \\ $$