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|(b+c,c+a,a+b),(c+a,a+b,b+c),(a+b,b+c,c+...

`|(b+c,c+a,a+b),(c+a,a+b,b+c),(a+b,b+c,c+a)|=`

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|(b+c,a+b,a),(c+a,b+c,b),(a+b,c+a,c)|=

|(a,b,c),(b,c,a),(c,a,b)|=

If |(a+b,b+c,c+a),(b+c,c+a,a+b),(c+a,a+b,b+c)|=k|(a,b,c),(b,c,a),(c,a,b)| then k=

|(b+c,a,1),(c+a,b,1),(a+b,c,1)|=

|(1+a,b,c),(a,1+b,c),(a,b,1+c)|=

|(a,b-c,c+b),(a+c,b,c-a),(a-b,b+a,c)|=

Show that |(b+c,c+a,a+b),(a+b,b+c,c+a)(a,b,c)|=a^(3)+b^(3)+c^(3)-3abc .

|(a-b,b-c,c-a),(b-c,c-a,a-b),(c-a,a-b,b-c)|=

|(a+b,c,c),(b,a+c,b),(a,a,b+c)|=