If #a+b+c=0#, then #(a^2+b^2+c^2)/(a^2b^2+b^2c^2+c^2a^2)=# ?

Question

5937 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-15T20:35:20

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

OR

#(a^2+b^2+c^2)/(a^2b^2+b^2c^2+c^2a^2)=4/(a^2+b^2+c^2)#

Here,

#a+b+c=0...to(1)#

Squaring both sides,

#(a+b+c)^2=0^2#

#=>a^2+b^2+c^2+2ab+2bc+2ca=0#

#color(red)(=>a^2+b^2+c^2=-2(ab+bc+ca)...to(2)#

Now,

#(ab+bc+ca)^2=a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc#

#(ab+bc+ca)^2=a^2b^2+b^2c^2+c^2a^2+2abc(b+c+a)#

#(ab+bc+ca)^2=a^2b^2+b^2c^2+c^2a^2+2abc(0)toFrom (1)#

#color(blue)((ab+bc+ca)^2=a^2b^2+b^2c^2+c^2a^2...to(3)#

So, from #color(red)((2)) andcolor(blue)( (3)#

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

OR

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

#=>(a^2+b^2+c^2)/(a^2b^2+b^2c^2+c^2a^2)=((-2)(-2))/(-2(ab+bc+ca)#

#=>(a^2+b^2+c^2)/(a^2b^2+b^2c^2+c^2a^2)=4/(a^2+b^2+c^2)tocolor(red)( From(2)#

Note:

If your question is :

#(a^2+b^2+c^2)^color(red)(2)/(a^2b^2+b^2c^2+c^2a^2)= ??, then,#

#(a^2+b^2+c^2)^2/(a^2b^2+b^2c^2+c^2a^2)=4#