x³+y³-1+3xy=(x+y)(x²-xy+y²)+3xy-1=(x+y)[(x+y)²-3xy]+3xy-1=(x+y)³-3xy(x+y)+3xy-1=[(x+y)-1][(x+y)²+(x+y)+1]-3xy[(x+y)-1]=[(x+y)-1][(x+y)²+(x+y)+1-3xy]=(x+y-1)(x&s...