" 12."ab+bc+a^(2)+ac

Factorise: 6ab - b ^(2) + 12 ac - 2bc

| [-bc, b ^ (2) + bc, c ^ (2) + bca ^ (2) + ac, -ac, c ^ (2) + aca ^ (2) + ab, b ^ (2) + ab, -ab (ab + bc + ac), is = 64. then

Find the square root of : 4 a ^(2) + 9b ^(2) + c ^(2) - 12 ab + 6 bc - 4 ac

Subtract : 2 ab + 5 bc - 7 ac from 5 ab - 2 bc - 2 ac + 10 abc

If ∣ -a a^2 ab ac ab -b^2 bc ac bc -c^2 | = ka^2b^2c^2 , then k is equal to

If the product abc =1, then the value of the determinant |{:( -a ^(2), ab, ac),( ba, -b ^(2), bc),(ac, bc,-c ^(2)):}| is

|[-bc, b^2+bc, c^2 +bc],[a^2+ac,-ac,c^2+ac],[a^2+ab,b^2+ab,-ab]|=64 . then (ab+bc+ac) is

Factorise the expressions : 6ab - b^2 + 12ac - 2bc .

Factorize: 6ab-b^(2)+12ac-2bc

Prove |[-bc, b^2+bc, c^2+bc] , [a^2+ac, -ac, c^2+ac] , [a^2+ab, b^2+ab, -ab]|=(ab+bc+ca)^2