Simplify the following:

(a−2b + 5c)(a−b)− (a−b−c)(2a + 3c) + (6a + b)(2c−3a−5b)

We have (a−2b + 5c)(a−b)− (a−b−c)(2a + 3c) + (6a + b)(2c−3a−5b)

Here, (a−2b + 5c)(a−b) = (a-b)(a-2b + 5c)(using Commutative law)

= a(a-2b + 5c)-b(a-2b + 5c)(using distributive law )

= a²-2ab + 5ac-ab + 2b²-5bc = a² + 2b²-3ab + 5ac-5bc

(a−b−c)(2a + 3c) = (2a + 3c)(a-b-c) (using Commutative law)

= 2a(a-b-c) + 3c(a-b-c) (using distributive law )

= 2a²-2ab-2ac + 3ac-3bc-3c²

= 2a² + ac-2ab-3bc-3c²

and (6a + b)(2c−3a−5b) = 6a(2c−3a−5b) + b(2c−3a−5b) = 12ac-18a²-30ab + 2bc-3ab-5b² = -18a² + 5b²-33ab + 2bc + 12ac

Therefore, a−2b + 5c)(a−b)− (a−b−c)(2a + 3c) + (6a + b)(2c−3a−5b) = a² + 2b²-3ab + 5ac-5bc-(2a² + ac-2ab-3bc-3c²) + ( -18a² + 5b²-33ab + 2bc + 12ac) =

a² + 2b²-3ab + 5ac-5bc-2a²-2ac + 2ab + 3bc + 3c²-18a² + 5b²-33ab + 2bc + 12ac =

= -19a² + 7b² + 3c²-34ab + 15ac