`|(a-b-c, 2a, 2a),(2b, b-c-a,2b),(2c,2c,c-a-b)| = (a + b + c)^(3)`.

|[1, a, a^2-b c],[1,b, b^2-c a],[1,c, c^2-a b]|=

The determinant |(b^2-ab,b-c,-ac),(ab-a^2,a-b,b^2-ab),(bc-ac,c-a,ab-a^2)| equals :

a. Minimize z =-3x+4y subject to constraints. x+2yle8 3x+2yle12 xge0, yge0 by graphical method. b. Prove that {:abs((1,a,a^2),(1,b,b^2),(1,c ,c^2)):} = (a - b)(b-c)(c-a)

Without expanding the determinant, prove that {:|( 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)) |:}

If a^2+b^2+c^2=0 and |(b^2+c^2,ab,ac),(ab,c^2+a^2,bc),(ac,bc,a^2+b^2)|=ka^2b^2c^2 , then the value of k is :

If alpha=|(a+b,b+c,c+a),(b+c,c+a,a+b),(c+a,a+b,b+c)|,beta=|(a,b,c),(b,c,a),(c,a,b)| , then

Prove that |(1,a^2,bc),(a,b^2,ca),(1,c^2,ab)|=(a-b)(b-c)(c-a)

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

Prove that {:|( a^(2) , bc, ac+c^(2)),( a^(2) +ab,b^(2) ,ac),( ab,b^(2) +bc,c^(2)) |:} =4a^(2) b^(2) c^(2)

Using the factor theorem it is found that b + c , c + a and a + b are three factors of the determine : |(-2a,a+b,a+c),(b+a,-2b,b+c),(c+a,c+b,-2c)| . The other factor in the value of the determine is :