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# If a+b+c=0, find the value of1/2a2+bc+1/2b2+ca+1/2c2+ab

Saurabh Koranglekar
one year ago
Dear student

Take out 1/2 common from required expression

0.5( a+b+c)^2

= 0.5*0*0 =0

Regards
Vikas TU
14149 Points
one year ago
Dear student
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
Hope this helps
Good Luck