a) 4y.(x-1) - (1-x) = 4y(x-1) + (x-1) = (4y+1)(x-1)
b) (x-3)3 + 3-x = (x-3)2(x-3) +(x-3) = [(x-3)2 +1](x-3) = (x2 -6x+10)(x-3)
\(\begin{array}{l} a,\ 4y.(x-1)-(1-x)\\ =3y.(x-1)+1.(x-1)\\ =(3y+1).(x-1)\\ b,\ (x-3)^3+3-x\\ =(x-3)^3-(x-3)\\ =(x-3).[(x-3)^2-1]\end{array}\)