|{:(-1,b,c),(a,-1,c),(a,b,-1):}|=(a+1)(b...

`|{:(-1,b,c),(a,-1,c),(a,b,-1):}|=(a+1)(b+1)(c+1)((a)/(a+1)+(b)/(b+1)+(c)/(c+1)-1)`

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|{:(1,a,b+c),(1,b,c+a),(1,c,a+b):}|=0

|{:(b+c,a,1),(c+a,b,1),(a+b,c,1):}|=0

|{:(1,1,1),(a^2,b^2,c^2),(a^3,b^3,c^3):}|=(b-c)(c-a)(a-b)(bc+ca+ab)

|{:(a-b,1,a),(b-c,1,b),(c-a,1,c):}|=|{:(a,1,b),(b,1,c),(c,1,a):}|

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|{:(1,b+c,b^2+c^2),(1,c+a,c^2+a^2),(1,a+b,a^2+b^2):}|=(b-c)(c-a)(a-b)

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If (1)/(a)+(1)/(c )=(1)/(2b-a)+(1)/(2b-c) , then

Let a,b,c be such that b(a+c) ne 0. If |{:(a,a+1,a-1),(-b,b+1,b-1),(c,c-1,c+1):}|+|{:(a+1,b+1,c-1),(a-1,b-1,c+1),((-1)^(n+2)a,(-1)^(n+1)b,(-1)^(n)c):}| =0 then the value of n is

CHHAYA PUBLICATION-DETERMINANT -EXERCISE 2B

|{:(1,b+c,b^2+c^2),(1,c+a,c^2+a^2),(1,a+b,a^2+b^2):}|=(b-c)(c-a)(a-b)
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|{:(a^2,ab,ac),(ab,b^2,bc),(ca,bc,c^2):}|=?
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|{:(-1,b,c),(a,-1,c),(a,b,-1):}|=(a+1)(b+1)(c+1)((a)/(a+1)+(b)/(b+1)+(...
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|{:(x^2+y^2+1,x^2+2y^2+3,x^2+3y^2+4),(y^2+2,2y^2+6,3y^2+8),(y^2+1,2y^2...
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|{:(" "3a," "-a+b,-a+c),(a-b," "3b," "c-b),(a-c," "b-c," "3c):}|...
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|{:(x,a,b),(a,x,b),(a,b,x):}|=(x-a)(x-b)(x+a+b)
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|{:(b+c,a-b,a),(c+a,b-c,b),(a+b,c-a,c):}|=3abc-a^3-b^3-c^3
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Prove |{:(1,a^2+bc,a^3),(1,b^2+ca,b^3),(1,c^2+ab,c^3):}|=-(a-b)(b-c)(c...
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|{:(a^2,a^2-(b-c)^2,bc),(b^2,b^2-(c-a)^2,ca),(c^2,c^2-(a-b)^2,ab):}|=(...
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|{:((a+1)(a+2),a+2,1),((a+2)(a+3),a+3,1),((a+3)(a+4),a+4,1):}|=-2
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Show that the following determinant is a perfact square : |{:(1,a,a^...
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if x+y+z = 0 then prove that |{:(x,y,z),(x^2,y^2,z^2),(y+z,z+x,x+y):}|...
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|{:(1,1,1),(x,y,z),(x^3,y^3,z^3):}|=0
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|{:(x," "4,-2),(4," "x,-2),(4,-2," "x):}|=0
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|{:(x,a,a),(a,x,b),(b,b,x):}|=0
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|{:(x+1," "2," "3),(" "1,x+1," "3),(" "3,-6,x+1):}|=0
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|{:(a-x," "b," "c),(" "b,c-x," "a),(" "c," "a,b-x):}|=0
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|{:(" "x," "c+x," "b+x),(" "c+x," "x," "a+x),(" "b+x," "a+x," ...
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|{:(2-x," "2," "3),(" "2,5-x," "6),(" "3," "4,10-x):}|=0
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solve:|{:(3+x,3-x,3-x),(3-x,3+x,3-x),(3-x,3-x,3+x):}|=0
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