Prove the following statement.

Asked by Priyanshu | 18th Jun, 2013, 01:36: PM

Expert Answer:

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

 

Answered by  | 18th Jun, 2013, 04:08: PM

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