=(x³+1)(x²+x+1)
=(x+1)(x²-+1)(x²+z+1)
I
(6) x+x+5x+4x²+4
=(z+z+z+4(x += ' +1)
=zz'+z²+1)+4(x+z²+1)
=(x*+4)(x*+z²+1) ₁?
={(x+4x²+4) — 4x³}{(x+2x²+1)-22)
= {(x²+2)² - (2x)³}{(x²+1)² −z ²}
={(x²+2)+2x}{(x²+2)−2)
×{(x²+1)+z}{(x²+ 1) − x}
=(x²+2x+2)(x²-2x+2)
x(x²+z+1)(x²-2+1)