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Solve-x-2-a-y-z-2-y-2-b-z-x-2-z-2-c-x-y-2-




Question Number 215356 by Abdullahrussell last updated on 04/Jan/25
 Solve:    x^2 =a+(y−z)^2     y^2 =b+(z−x)^2     z^2 =c+(x−y)^2
Solve:x2=a+(yz)2y2=b+(zx)2z2=c+(xy)2
Answered by mr W last updated on 04/Jan/25
(x+y−z)(x−y+z)=a   ...(i)  (−x+y+z)(x+y−z)=b   ...(ii)  (x−y+z)(−x+y+z)=c   ...(iii)  ⇒(−x+y+z)^2 (x−y+z)^2 (x+y−z)^2 =abc  assume abc≥0, otherwise no solution.  (−x+y+z)(x−y+z)(x+y−z)=±(√(abc))  ⇒−x+y+z=±((√(abc))/a)   ...(I)  ⇒x−y+z=±((√(abc))/b)   ...(II)  ⇒x+y−z=±((√(abc))/c)   ...(III)  (II)+(III):  ⇒2x=±((1/b)+(1/c))(√(abc))  x=±(1/2)((1/b)+(1/c))(√(abc))  similarly  y=±(1/2)((1/c)+(1/a))(√(abc))  z=±(1/2)((1/a)+(1/b))(√(abc))
(x+yz)(xy+z)=a(i)(x+y+z)(x+yz)=b(ii)(xy+z)(x+y+z)=c(iii)(x+y+z)2(xy+z)2(x+yz)2=abcassumeabc0,otherwisenosolution.(x+y+z)(xy+z)(x+yz)=±abcx+y+z=±abca(I)xy+z=±abcb(II)x+yz=±abcc(III)(II)+(III):2x=±(1b+1c)abcx=±12(1b+1c)abcsimilarlyy=±12(1c+1a)abcz=±12(1a+1b)abc

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