If # a+b+c=0# then prove that #1/(2a ^2 +bc) +1/(2b^2+ca)+1/(2c^2+ab)=0# ?

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Answer 1 · 2018-02-26T09:00:44

Please see below.

As #a+b+c=0#, #a=-(b+c)# or #a^2=b^2+c^2+2bc#

and #2a^2+bc=a^2+a^2+bc#

= #a^2+a(-b-c)+bc#

=#a(a-b)-c(a-b)#

= #(a-c)(a-b)#

and similarly #2b^+ca)=(b-c)(b-a)# and #2c^2+ab=(c-a)(c-b)#

Hence #1/(2a^2 +bc) +1/(2b^2+ca)+1/(2c^2+ab)#

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

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

= #0#

Answer 2 · 2018-02-26T11:14:16

Please see a Proof in the Explanation.

In the given Expression (Exp.), we have #3" variables"#.

We try to reduce the no. of variables from #3# to #2#

by using the given condition that #a+b+c=0#.

#a+b+c=0 rArr c=-a-b#.

#:. 2a^2+bc=2a^2+b(-a-b)=2a^2-ab-b^2#,

#=(a-b)(2a+b)#.

Similarly, #2b^2+ca=2b^2+a(-a-b)=2b^2-ab-a^2#,

#=(b-a)(2b+a)#.

Finally, #2c^2+ab=2(-a-b)^2+ab=2a^2+5ab+2b^2#,

#=(a+2b)(2a+b)#.

For ease of writing, let,

#(a-b)=x, 2a+b=y, and 2b+a=z#.

#:."The Exp.="1/(xy)+1/{(-x)z}+1/(yz)#,

#=(z-y+x)/(xyz)#,

#={(2b+a)-(2a+b)+(a-b)}/(xyz)#.

#rArr"The Exp.="0,# as desired!