设(axb)·c=2,则[(a+b)x(b+c)]

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[（a+b）×(b+c)]·(c+a)
=(a×b+b×b+a×c+bxc)·(c+a)
=(a×b+0+a×c+bxc)(c+a) [注意：b×b=0]
=(a×b)·c+ ( b×c )·a [注意：(a×c)·c=0,【∵a×c⊥c】,同样0=(b×c)·c=(a×b)·a=(a×c)·a]
=2(a×b)·c=2×2=4