prakash ojha
Last Activity: 5 Years ago
2a²+bc=2(b+c)²+bc
=2(b²+2bc+c²)+bc
=2b²+4bc+2c²+bc
=2b²+5bc+2c²
=(2b+c)(b+2c)
=(b+b+c)(b+c+c)
=(b-a)(-a+c) [since a+b+c=0]
=(b-a)(c-a)
Similarly,
2b²+ca=(c-b)(a-b)
and
2c²+ab=(a-c)(b-c)
Now,
1/2a²+bc + 1/2b²+ca + 1/2c²+ab
=1/(b-a)(c-a) + 1/(c-b)(a-b) + 1/(a-c)(b-c)
=1/(a-b)(a-c) -1/(b-c)(a-b) + 1/(a-c)(b-c)
={b-c-(a-c)+a-b)}/(a-b)(a-c)(b-c)
={b-c-a+c+a-b}/(a-b)(