Discussion Forum

 If   $[a\times b .b\times c. c\times a]=\lambda[abc]^{2}$  then λ  is equal to

Answer:

Explanation:

 Use the following formula to simplify

   a x (bx c)=(a.c)b-(a.b)c

               [abc]=[bca]=[cab]

 and [a a b]=[a b b]=[a c c]=0

Now, [ a x ba x c c x a]

      = a xb.(( b xc ) x (c xa))

      = a x b.(k x (c x a))

 Let k= b x c

        = a x  b.((k.a) c-(k.c).a)

         =  (a x b).( ( b x c. a) c -(b x c. c) a)

  = ( a x b). ([b c a ] c)-0

                               [ $\because$   [  b x c.c ]=0]

 = (a x b).c [ b c a]

  = [ a b c ] [ b c a]= [ a b c]²

                                  [   $\because$  [a b c]=[b c a] ]

 But given,

  [ a x b b x c  c xa]= λ  [ a b c]²

So  [a b c]²  = λ  [ a b c ]²

$\Rightarrow$                 $\lambda$  =1