गुणनखंड निकाले `5a^(2) + 10ab + 15ac`

a ^(2) + bc + ab + ac= ?

Divide : ( 20a^(2) - 15 ab + 45a ) by (5a)

| [-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

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

The determinant |(a^(2)+10,ab, ac),(ab, b^(2)+10, bc),(ac, bc, c^(2)+10)| is