if a+b+c=0 then prove that a2/(2a2+bc)+b2/(2b2+ac)+c2/(2c2+ab)=1
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Answered by
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You can write your expression as:
1/(2a2+bc/a2)+1/(2+ca/b2)+1/(2+ab/c2)=... which reduces to:
a2/bc + b2/ac + c2/ab = 3. If you now plug a = -(b+c) you will get:
(b+c)^2/bc - (b^2/c + c^2/b)/(b+c) = 3, which reduces to
[(b+c)^3 - b^3 - c^3]/[bc*(b+c)] = 3*bc*(b+c)/[bc*(b+c)] which is
indeed equal 3 as needed.
1/(2a2+bc/a2)+1/(2+ca/b2)+1/(2+ab/c2)=... which reduces to:
a2/bc + b2/ac + c2/ab = 3. If you now plug a = -(b+c) you will get:
(b+c)^2/bc - (b^2/c + c^2/b)/(b+c) = 3, which reduces to
[(b+c)^3 - b^3 - c^3]/[bc*(b+c)] = 3*bc*(b+c)/[bc*(b+c)] which is
indeed equal 3 as needed.
Answered by
2
Answer:
a+c+b = 0
a = -b-c
ca = -bc - c²
2b² + ca = b² - c² + b² - bc
= - (b-c)(a-b)
lly, 2a² + bc = -(a-b)(c-a) and 2c² + ab = -(c-a)(b-c)
Now LHS = -a²/(a-b)(c-a) - b²/(b-c)(a-b) - c²/ (c-a) (b-c)
now by LCM and multiplication
= -{a²b-a²c+b²c-ab²+ac²-bc² / - (a²b-a²c+b²c -ab² + ac² - bc²)]
= 1 = RHS
Hence Proved.
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