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Delta=|[1,a,bc],[1,b,ca],[1,c,ab]|...

Delta=|[1,a,bc],[1,b,ca],[1,c,ab]|

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Evaluate : triangle = |[1,a,bc],[1,b,ca],[1,c,ab]|

Evaluate: |[1,a,bc],[1,b,ca],[1,c,ab]|

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

Delta=det[[1,a,bc1,b,ca1,c,ab]]

Show without expanding at any stage that: [1/a, a,bc],[1/b,b,ca],[1/c, c, ab]|=0

If Delta_(1)=|(1,a,bc),(1,b,ca),(1,c,ab)|,Delta_(2)=|(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2))| then

Prove that . |[1,a,a^2-bc],[1,b,b^2-ca],[1,c,c^2-ab]|=0

Evaluate Delta = {:[ (1,a,bc),( 1,b,ca),(1,c,ab)]:}

Evaluate Delta = {:[ (1,a,bc),( 1,b,ca),(1,c,ab)]:}

Factorise : |(1,a,bc),(1,b,ca),(1,c,ab)|