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Question
Prove that `(bar"a" + 2bar"b" - bar"c"). [(bar"a" - bar"b") xx (bar"a" - bar"b" - bar"c")] = 3 [bar"a" bar"b" bar"c"]`.
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Solution
`(bar"a" + 2bar"b" - bar"c"). [(bar"a" - bar"b") xx (bar"a" - bar"b" - bar"c")] = 3 [bar"a" bar"b" bar"c"]`
`= (bar"a" + 2bar"b" - bar"c").(bar"a"xxbar"a" - bar"a" xx bar"b" - bar"a" xx bar"c" - bar"b" xx bar"a" + bar"b" xx bar"b" + bar"b" xx bar"c")`
`= (bar"a" + 2bar"b" - bar"c").(bar"0" - bar"a" xx bar"b" - bar"a" xx bar"c" + bar"a" xx bar"b" + bar"0" + bar"b" xx bar"c")`
`= (bar"a" + 2bar"b" - bar"c"). (bar"c" xx bar"a" + bar"b" xx bar"c")`
`= bar"a".(bar"c" xx bar"a") + bar"a".(bar"b"xxbar"c") + 2bar"b".(bar"c" xx bar"a") + 2bar"b".(bar"b"xxbar"c") - bar"c".(bar"c" xx bar"a") - bar"c".(bar"b" xx bar"c")`
`= 0 + bar"a".(bar"b" xx bar"c") + 2bar"b".(bar"c" xx bar"a") + 2 xx 0 - 0 - 0`
`= [bar"a" bar"b" bar"c"] + 2[bar"b" bar"c" bar"a"]`
`= [bar"a" bar"b" bar"c"] + 2[bar"a" bar"b" bar"c"]`
= `3[bar"a" bar"b" bar"c"]`
Notes
The question in the textbook has a printing mistake, it should be `3 [bar"a" bar"b" bar"c"]` not 3 `[bar"a" - bar"b" - bar"c"]`.
