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" 14."|[1,a,bc],[1,b,ca],[1,c,ab]|=(a-b)...

" 14."|[1,a,bc],[1,b,ca],[1,c,ab]|=(a-b)(b-c)(c-a)

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|[1,1,1],[a,b,c],[bc,ca,ab]|=(a-b)(b-c)(c-a)

Using properties of determinants prove the following. abs[[1,a,bc],[1,b,ca],[1,c,ab]]=(a-b)(b-c)(c-a)

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