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|[(b+c)^2,ab,ca],[ab,(a+c)^2,bc],[ac,bc,...

`|[(b+c)^2,ab,ca],[ab,(a+c)^2,bc],[ac,bc,(a+b)^2]|=2abc(a+b+c)^3`

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Prove that |[(a+b)^(2),ca,bc],[ca,(b+c)^(2),ab],[bc,ab,(c+a)^(2)]|=2abc(a+b+c)^(3)

(b + c) ^ (2), ab, caab, (a + c) ^ (2), bcac, bc, (a + b) ^ (2)] | = 2abc (a + b + c) ^ ( 3)

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If A=[[a^2,ab,ac],[ab,b^2,bc],[ac,bc,c^2]] and a^2+b^2+c^2=1, then A^2

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Prove that identities: |[-bc,b^2+bc,c^2+bc],[a^2+ac,-ac,c^2+ac],[a^2+ab,b^2+ab,-ab]|=(a b+b c+a c)^3

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