ICSE Class 10 Answered
Prove the following statement.
Asked by Priyanshu | 18 Jun, 2013, 01:36: PM
Expert Answer
Given :If (x+y)
(3a-b)=(y+z)(3b-c)=(z+x)(3c-a)
(3a-b)=(y+z)(3b-c)=(z+x)(3c-a)To prove :(x+y+z)(a+b+c)=(ax+by+cz)(a2 + b2+ c2 )
(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
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hence proved
Answered by | 18 Jun, 2013, 04:08: PM
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