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|{:(1,a,b+c),(1,b,c+a),(1,c,a+b):}|=0...

`|{:(1,a,b+c),(1,b,c+a),(1,c,a+b):}|=0`

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The correct Answer is:
`=(a+b+c)xx0[because` 1 st and 2nd columns are identical ] =0
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|{:(1,a,bc),(1,b,ca),(1,c,ab):}|=|{:(1,a,a^2),(1,b,b^2),(1,c,c^2):}|

Without expanding prove that, |{:(1,bc,b+c),(1,ca,c+a),(1,ab,a+b):}|=|{:(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2)):}|

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prove that : |{:(1,a,b),(-a,1,c),(-b,-c,1):}|=1+a^2+b^2+c^2

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If |{:(a,b,1),(b,c,1),(c,a,1):}|=2010 and if |{:(c-a,c-b,ab),(a-b,a-c,bc),(b-c,b-a,ca):}|-|{:(c-a,c-b,c^(2)),(a-b,a-c,a^(2)),(b-c,b-a,b^(2)):}|=p , then the number of positive divisors of p is

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