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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 ]

(x+y+z)2=x3+y3+z2−3xyz(x+y+z)2=(x+y+z)(x2+y2+z2−(xy+xz+yz))[1]★x+y+z=0⇔x=−(y+z)∧(y,z)∈R2[2]x+y+z=x2+y2+z2−(xy+xz+yz)y=px∧z=qx(p+q+1)x=(p2−pq+q2−p−q+1)x2[2.1]★x=0⇒y=0∧z=0[alreadyincludedabove][2.2]p+q+1=(p2−pq+q2−p−q+1)x★x=p+q+1p2−pq+q2−p−q+1∧y=px∧z=qx∧(p,q)∈R2[p2−pq+q2−p−q+1≠0∀(p,q)∈R2]

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