Given `Delta=[(a+b,b+c,c+a),(b+c,c+a,a+b),(c+a,a+b,b+c)]` Express given determinant as the sum of 8 determinants.

Evaluate Delta=|[1, a, bc],[ 1, b, c a],[ 1, c, a b]|

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

prove |[1, b c, a(b+c)],[ 1, c a, b(c+a)],[ 1, a b, c(a+b)]|=0

|{:(a+x, b, c), (a, b+y, c), (a, b, c+z):}| is equal to :

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

If |(b+c,c+a,a+b),(a+b,b+c,c+a),(c+a,a+b,b+c)|=k|(a,b,c),(c,a,b),(b,c,a)| then the value of k is

If (a)/(b+c),(b)/(c+a),(c)/(a+b) are in AP, then

The determinant Delta=|(a,b,ax+b),(b,c,bx+c),(ax+b,bx+c,c)| is equal to zero, if a)a,b,c are in A.P. b)a,b,c are in G.P. c)a,b,c are in H.P. d) ax^(2) + bx + c = 0

Consider the points (b,c+a),(c,a+b) and (a,b+c). Are the given points collinear? Why?