Vishal Dagur
Last Activity: 3 Years ago
Given A2+b2+C2= -(ab+bc+ca) then find 3a-2b+5c
So, multiple both side 2
2a2 +2b2 +2c2 = -2ab-2bc-2ca
We right -- 2a2 = a2 +a2 ,2b2=b2 +b2 ,2c2=c2+c2
So,
a2+a2+b2+b2+c2+c2+2ab+2bc+2ca=0
This make formula
(A+B)2+(B+C)2+(C+A)2=0
(A+B)2=0 ,A+B =0 ,so A= -B
(B+C)2=0 , B=-C
(C+A)2=0 , C= -A
Now
3a-2b+5c.
Put a =-B and c=-A
-3b-2b+5(-A). (We know a=-B so -A =b)
-3b-2b+5b
-5b+5b = 0.
So there answer is zero