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|{:(1,a,bc),(1,b,ca),(1,c,ab):}|=|{:(1,a...

`|{:(1,a,bc),(1,b,ca),(1,c,ab):}|=|{:(1,a,a^2),(1,b,b^2),(1,c,c^2):}|`

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The correct Answer is:
[interchanging 1 st and 2 nd columns].
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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)):}|

|{:(a,a^2,bc),(b,b^2,ca),(c,c^2,ab):}|=|{:(1,a^2,a^3),(1,b^2,b^3),(1,c^2,c^3):}|

|{:(a-b,1,a),(b-c,1,b),(c-a,1,c):}|=|{:(a,1,b),(b,1,c),(c,1,a):}|

|{:(1,1,1),(a,b,c),(bc,ca,ab):}|=(a-b)(b-c)(c-a)

|{:(1,bc,a(b+c)),(1,ca,b(c+a)),(1,ab,c(a+b)):}|=0

Prove without expanding |[a,a^2,bc],[b,b^2,ca],[c,c^2,ab]|=|[1,a^2,a^3],[1,b^2,b^3],[1,c^2,c^3]|

|{:(1,bc,bc(b+c)),(1,ca,ca(c+a)),(1,ab,ab(a+b)):}|=0

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

Prove |{:(1,a^2+bc,a^3),(1,b^2+ca,b^3),(1,c^2+ab,c^3):}|=-(a-b)(b-c)(c-a)(a^2+b^2+c^2)

prove that , |{:(a,a^2,a^3+bc),(b,b^2,b^3+ca),(c,c^2,c^3+ab):}|=(a-b)(b-c)(c-a)(abc+bc+ca+ab)