Question

if bcx =acy =abz then show that ax+by/a^2+b^2 = by+cz/b^2+c^2 = ax+cz/a^2+c^2​

Answer 1

Answer:

We add the additional constraint that a, b, c are nonzero, for otherwise the statement is not true.  For example, a=x=0, b=c=y=1, z=2 satisfies bcx=acy=abz and yet the other values are not equal, being 1, 3/2 and 2.

Since a, b, c are not zero, the given constraint can be rewritten:

bcx = acy => bx = ay

acy = abz => cy = bz

abz = bcx => az = cx

Therefore

ab ( bx - ay ) + bc ( cy - bz ) + ca ( cx - az ) = 0

=> ab²x + bc²y + ac²x = a²by + b²cz + a²cz

=> ab²x + bc²y + ac²x + b³y = a²by + b²cz + a²cz + b³y

=> ( ax + by ) ( b² + c² ) = ( by + cz ) ( a² + b² )

=> ( ax + by ) / ( a² + b² ) = ( by + cz ) / ( b² + c² )

The other equations follow similarly.