|{:(a^2,a^2-(b-c)^2,bc),(b^2,b^2-(c-a)^2...

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

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

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)

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)

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

Using properties of determinants prove that, |{:((b+c)^(2),a^(2),a^(2)),(b^(2),(c+a)^(2),b^(2)),(c^(2),c^(2),(a+b)^(2)):}|=2abc(a+b+c)^(3)

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

|{:(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):}|

Show that: abs(((b+c)^2,a^2,bc) , ((c+a)^2,b^2,ca) , ((a+b)^2,c^2,ab))=(a^2+b^2+c^2)(a+b+c)(b-c)(c-a)(a-b)

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