(ii)|[1,1,1],[a,b,c],[a',b',c^(4)]|=(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)

1,1,1a,b,ca^(3),b^(3),c^(3)]|=(a-b)(b-c)(c-a)(a+b+c)

Show that |(1,1,1), (a,b,c),(a^2,b^2,c^2)|=(a-b)(b-c)(c-a)

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)):}

By using properties of determinants. Show that: (i) |1a a^2 1bb^2 1cc^2|=(a-b)(b-c)(c-a) (ii) |1 1 1a b c a^3b^3c^3|=(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

Prove that |(1,1,1),(bc,ca,ab),(b+c, c+a, a+b)| = (a-b)(b-c)(c-a)

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 : (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+a,b,c),(a,1+b,c),(a,b,1+c)|=