Simplify the following:

(x−2y)(y−3x) + (x + y)(x−3y)−(y−3x)(4x−5y)

Here we have (x−2y)(y−3x) + (x + y)(x−3y)−(y−3x)(4x−5y)

We will use the distributive law as follows:

Now (x−2y)(y−3x) = x(y-3x)-2y(y-3x) = x.y-3x.x-2y.y + 6x.y = xy-3x²-2y² + 6xy

(x + y)(x−3y) = x(x-3y) + y(x-3y) = x.x-3x.y + x.y-y.3y = x²-3xy + xy-3y² = x²-2xy-3y²

(y−3x)(4x−5y) = y(4x-5y)-3x(4x-5y) = 4x.y-5y.y-12x.x + 15x.y = 4xy-5y²-12x² + 15xy

Therefore , (x−2y)(y−3x) + (x + y)(x−3y)−(y−3x)(4x−5y) = xy-3x²-2y² + 6xy + x²-2xy-3y²-4xy-5y²-12x² + 15xy = 16xy-14x²-10y²

Hence, (x−2y)(y−3x) + (x + y)(x−3y)−(y−3x)(4x−5y) = -14x²-10y² + 16xy