[" 21."|[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^(3),b^(3),c^(3)]|=(a-b)(b-c)(c-a)(a+b+c)

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)

By using properties of determinants, show that : |[1,1,1],[a,b,c],[a^3,b^3,c^3]| = (a-b)(b-c)(c-a)(a+b+c)

Prove that |(1,1,1),(a,b,c),(a^(3),b^(3),c^(3))|=(a-b)(b-c)(c-a)(a+b+c)

Using the property of determinants and without expanding prove that abs([1,1,1],[a,b,c],[a^3,b^3,c^3])=(a-b)(b-c)(c-a)(a+b+c)

Using properties of determinants prove the following. abs[[1,1,1],[a,b,c],[a^3,b^3,c^3]]=(a-b)(b-c)(c-a)(a+b+c)

Prove that: (i) |{:(,1,1,1),(,a,b,c),(,a^(3),b^(3),c^(3)):}|=(a-b)(b-c)(c-a)(a+b+c)

Prove that |{:(1, 1, 1),(a, b, c),(a^(3), b^(3), c^(3)):}|=(a-b)(b-c)(c-a)(a+b+c)

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)):}

If |(1,1,1),(a,b,c),(a^(3),b^(3),c^(3))| = (a - b) (b - c) (c - a) (a + b + c) , where a,b,c are all different, then the determinant |(1,1,1),((x-a)^(2),(x-b)^(2),(x-c)^(2)),((x-b)(x-c),(x-c)(x-a),(x-a)(x-b))| vanishes when a)a + b + c = 0 b) x = (1)/(3) (a + b + c) c) x = (1)/(2) (a + b + c) d) x = a + b + c