(i) (10ab+22bc−13a−14b−16c−24)−(5ab−6a+15c−15+4ac)
10ab+22bc−13a−14b−16c−24−5ab+6a−15c+15−4ac
=5ab+22bc−7a−14b−31c−9−4ac
(ii)(50−20b2c2−4bc−ac−5ab)−(32+10b2c2−21ab+5a2b−10b2c)
50−20b2c2−4bc−ac−5ab−32−10b2c2+21ab−5a2b+10b2c
=18−30b2c2−4bc−ac+16ab−5a2b+10b2c