Question

Prove that [a+b,b+c,c+a]=2[a,b,c]

Answer 1

hey there...

here is ur answer

if we add all those terms then if we get

$a + b + b + c + c + a$

we get

$2a + 2b + 2c$

if we take common in these we get..

$2(a + b + c)$

hence proved.

hope this helps!

Answer 2

[ a+b b+c c+a ] = ((a+b). (b+c))×(c+a) expanding
 = ((a.b)+(a.c)+(b.b)+(b.c))×(c+a) expanding
 = ((a.b)×c)+((a.c)×c+((b.b)×c)+((b.c)×c) + ((a.b)×a)+((a.c)×a)+((b.b)×a)+((b.c)×a)
 = [ a b c ]+[ a c c ]+[ b b c ]+[ a c c ] + [ a b a ]+[ a c a ]+[ b b a ]+[ b c a ] 
 = [ a b c ] + [ b c a ]

since, [ a b b ] = 0 i.e. if same vectors are in the box product it's value is zero

= [ a b c ]+[ a b c ]