Prove |{:(1,a^2+bc,a^3),(1,b^2+ca,b^3),(...

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

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

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

Without expanding the determinant, prove that |{:(1,a,a^2),(1,b,b^2),(1,c,c^2):}|=(a-b)(b-c)(c-a) .

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

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

Using determinant : show that the area of the triangle with vertices at (a^2,a^3),(b^2,b^3)and (c^2,c^3) is 1/2(a-b)(b-c)(c-a)(ab+bc+ca) square unit.

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,bc,bc(b+c)),(1,ca,ca(c+a)),(1,ab,ab(a+b)):}|=0

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

CHHAYA PUBLICATION-DETERMINANT -EXERCISE 2B

|{:(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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|{:(x-2,2x-3,3x-4),(x-4,2x-9,3x-16),(x-8,2x-27,3x-64):}|=0
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|{:(0,x-a,x-b),(x+a,0,x-c),(x+b,x+c,0):}|=0[where" "a neb nec]
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Solve: |{:(3,5,x),(7,x,7),(x,5,3):}|=0
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|{:(x+a," "b," "c),(" "a,x+b," "c),(" "a," "b,x+c):}|=0
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|{:(3x-8," "3," "3),(" "3,3x-8," "3),(" "3," "3,3x-8):}|=0
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