a) ( 2x +3)2 + (2x-3)2 + (2x+3)(4x-6) + xy
= (2x+3)2 + 2(2x+3)(2x-3) + xy
= \([\) (2x+3) + (2x-3) \(]\)2 + xy
= (4x)2 + xy = 16x2 + xy = x(16 + y)
b) x2 + x - y2 + y
= (x2 - y2 ) + ( x + y )
= (x+y)(x-y) + (x+y)
= (x+y)(x-y+1)
c) 3x2 + 3y2 - 6xy - 12
= 3(x2 + y2 - 2xy - 4)
= 3[ (x-y)2 -22 ] = 3(x-y-2)(x-y+2)
d) x3 -x + 3x2y + 3xy2 -y + y3
= ( x3 + 3x2y + 3xy2 + y3 ) - (x + y)
= (x+y)3 - (x+y)
= (x+y)[ (x+y)2 - 1 ] = (x+y)(x+y-1)(x+y+1)
e) 2018x2 - 2019x + 1 = 0
=> 2018x2 - 2018x - x + 1 = 0
=> 2018x(x-1) - (x-1) = 0
=> (x-1)(2018x-1) = 0
=> \(\left[{}\begin{matrix}x-1=0\\2018x-1=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2018}\end{matrix}\right.\)
