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(a)|[1,1,1],[a,b,c=(a-b)(b-c)(c-a)(a+b+c...

(a)|[1,1,1],[a,b,c=(a-b)(b-c)(c-a)(a+b+c)],[a^(2),b^(2),c]|

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Match the following from List - I to List - II {:("List-I","List-II"),((I)|{:(1,1,1),(a,b,c),(bc,ca,ab):}|=,(a)(a-b)(b-c)(c-a)),((II)|{:(a,b,c),(a^(2),b^(2),c^(2)),(a^(3),b^(3),c^(3)):}|=,(b)(a-b)(b-c)(c-a)abc),((III)|{:(1,1,1),(a,b,c),(a^(3),b^(3),c^(3)):}|=,(c)(a-b)(b-c)(c-a)(a+b+c)):}

([1,1,1a,b,ca^(2),b^(2),c^(2)])=(a-b)(b-c)(c-a)

|(1,a^(2),a^(4)),(1,b^(2),b^(4)),(1,c^(2),c^(4))|=(a+b)(b+c)(c+a)|(1,1,1),(a,b,c),(a^(2),b^(2),c^(2))|

If f(a,b) =(f(b)-f(a))/(b-a) and f(a,b,c)=(f(b,c)-f(a,b))/(c-a) prove that f(a,b,c)=|{:(f(a),f(b),f(c)),(1,1,1),(a,b,c):}|-:|{:(1,1,1),(a,b,c),(a^(2),b^(2),c^(2)):}| .

If f(a,b) =(f(b)-f(a))/(b-a) and f(a,b,c)=(f(b,c)-f(a,b))/(c-a) prove that f(a,b,c)=|{:(f(a),f(b),f(c)),(1,1,1),(a,b,c):}|-:|{:(1,1,1),(a,b,c),(a^(2),b^(2),c^(2)):}| .

If f(a,b) =(f(b)-f(a))/(b-a) and f(a,b,c)=(f(b,c)-f(a,b))/(c-a) prove that f(a,b,c)=|{:(f(a),f(b),f(c)),(1,1,1),(a,b,c):}|-:|{:(1,1,1),(a,b,c),(a^(2),b^(2),c^(2)):}| .

By using properties of determinants. Show that: (i) |[1,a, a^2],[ 1,b,b^2],[ 1,c,c^2]|=(a-b)(b-c)(c-a) (ii) |[1, 1, 1],[a, b, c],[ a^3,b^3,c^3]|=(a-b)(b-c)(c-a)(a+b+c)

1,1,1a,b,ca^(2),b^(2),c^(2)]|=(a-b)(b-c)(c-a)

Prove: |[1,a^2+bc, a^3],[ 1,b^2+c a, b^3],[ 1,c^2+a b, c^3]|=-(a-b)(b-c)(c-a)(a^2+b^2+c^2)

Prove that: |[1, 1, 1],[a, b, c],[a^2, b^2, c^2]|=(a-b)(b-c)(c-a)