Сократите дроби: 1-с 2/1-с 3 8t3+125/4t2-25 b2-4/b3-8 16z2-9/27-64z3

1-с²/1-с³=(1-с) (1+ с) / (1-с) (1+ с + с²)=1+ с/1+ с + с²8t³+125/4t²-25=(2t+5) (2t²-10t+25) / (2t+5) (2t-5)=2t²-10t+25/2t-5b²-4/b³-8=(b-2) (b+2) / (b-2) (b²+2b+4)=b+2/b²+2b+416z²-9/27-64z³=(4z-3) (4z+3) / (3-4z) (9+12z+16z²)=- (3-4z) (4z+3) / (3-4z) (9+12z+16z²)=-4z-3/9+12z+16z²