Question

If a+b+c=0 then prove that 1/(2a2+bc1/(2b2+ac1/(2c2+ab) =0

Answer 1

Answer:

Step-by-step explanation:

∴ a + b + c = 0

∴ a = -(b + c), b = -(c + a) , c = -(a + b)

Now, (2a² + bc ) = (a² + a² + bc)

= {a² + a(-b - c) + bc}

= a² - ab - ac + bc = a(a - b) -c(a - c) = (a - c)(a - b) ------------(1)

similarly, (2b² + ca) = (b - c)(b - a) -------------(2)

(2c² + ab) = (c - a)(c - b) --------------(3)

now, LHS = 1/(2a² + bc ) + 1/(2b² + ca) + 1/(2c² + ab)

= 1/(a - b)(a - c) + 1/(b - c)(b - a) + 1/(c - a)(c - b)

= - 1/(a- b)(c - a) -1/(b - c)(a - b) - 1/(c - a)(b - c)

= -[(b -c ) + (c - a) + (a - b)]/(a - b)(b - c)(c - a)]

= 0 = RHS

Hence, proved

Answer 2

Answer:

hope this will help you

Step-by-step explanation:

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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