Prove the following statement.

Given  :If  (x+y)(3a-b)=(y+z)(3b-c)=(z+x)(3c-a)

To prove :(x+y+z)(a+b+c)=(ax+by+cz)(a² + b²+ c² )

You have to apply the rule of identities.

a1/b1 = a2/b2 = a3/b3 
= (a1+a2+a3)/(b1+b2+b3) ---- 1
= (a1-a2+a3)/(b1-b2+b3) ---- 2
= (a2-a3+a1)/(b2-b3+b1) ---- 3
= (a3-a1+a2)/(b3-b1+b2) ---- 4


apply to the given equations:

=(x+y) /(3a-b) = (y+z)/3b-c) = (z+x)/3c-a)

=(x+y+y+z+z+x) /(3a-b+3b-c +3c-a)......{using 1}

=(x+y+z)/ (a+b+c)

=(x+y-y-z+z+x ) /(3a-b-3b+c+3c-a)......{using 2 }

=2x/(2a-4b+4c)

=x/(a-2b+2c)

=ax/(a²-2ab+2ac)  ---- result A  {multiply n divide by a }

 

=(y+z-z-x+x+y) /(3b-c-3c+a+3a-b)   {using 3}

=2y/(2b-4c+4a)

=y/(b-2c+2a)

 

=by/(b²-2bc+2ab)....result B

=  ---- by(4)
= 
= 
=  ---- result C

=  -------- by adding results A,B,C

=  = RHS

hence proved