We will use the properties of the scalar triple product to expand the left side of the equation:
[a+b,b+c,c+a]=(a+b)⋅((b+c)×(c+a)) First, expand the cross product:
(b+c)×(c+a)=b×c+b×a+c×c+c×a Since c×c=0, we have: (b+c)×(c+a)=b×c+b×a+c×a Now, take the dot product with (a+b): (a+b)⋅(b×c+b×a+c×a)=a⋅(b×c+b×a+c×a)+b⋅(b×c+b×a+c×a) Distribute the dot product:
a⋅(b×c)+a⋅(b×a)+a⋅(c×a)+b⋅(b×c)+b⋅(b×a)+b⋅(c×a) Recall that if two vectors in a scalar triple product are the same, the scalar triple product is zero.
Therefore, a⋅(b×a)=0, a⋅(c×a)=0, b⋅(b×c)=0, and b⋅(b×a)=0. We are left with:
a⋅(b×c)+b⋅(c×a) We know that a⋅(b×c)=[a,b,c] and b⋅(c×a)=[b,c,a]. Also, the scalar triple product is invariant under cyclic permutations: [a,b,c]=[b,c,a]=[c,a,b]. Therefore, b⋅(c×a)=[b,c,a]=[a,b,c]. So we have:
[a,b,c]+[a,b,c]=2[a,b,c] Thus, [a+b,b+c,c+a]=2[a,b,c].