" 8.10prove "|[x+a,b,c],[a,x+b,c],[a,b,x+c]|=x^(2)(x+a+b+c)

S.T abs[[x+a,b,c],[a,x+b,c],[a,b,x+c]]=x^2(x+a+b+c)

Using the properties of determinants, prove that : |[[x+a,b,c],[a,x+b,c],[a,b,x+c]]| = x^2(x+a+b+c)

Prove that : |{:(x+a,b,c),(a,x+b,c),(a,b,x+c):}|=x^(2)(x+a+b+c)

Prove that : |{:(x+a,b,c),(a,x+b,c),(a,b,x+c):}|=x^(2)(x+a+b+c)

The determinant |[y^2,-x y, x^2],[a, b, c],[ a ', b ', c ']| is equal to a. |[b x+a y, c x+b y],[ b^(prime)x+a ' y, c^(prime)x+b ' y]| b. |[a x+b y, b x+c y],[ a^(prime)x+b ' y, b ' x+c ' y]| c. |[b x+c y, a x+b y],[ b^(prime)x+c ' y, a^(prime)x+b ' y]| d. |[a x+b y, b x+c y],[ a^(prime)x+b ' y, b^(prime)x+c ' y]|

if x^(3) =1,prove : |{:(a,b,c),(b,c,a),(c,a,b):}|=(a+bx+cx^2)|{:(1,b,c),(x^2,c,a),(x,a,b):}|

If a+b+c=0 and |[a-x,c,b],[c,b-x,a],[b,a,c-x]|=0 then x=

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 a ,b ,c are different, then the value of |[0,x^2-a, x^3-b],[ x^2+a,0,x^2+c],[ x^4+b, x-c,0]| is a. c b. a c. b d. 0

a+b+c=0, solve for x : [[a-x,c,b],[c,b-x,a],[b,a,c-x]]=0