Question 5(d)
Substract: 3a(a+b+c)−2b(a−b+c)from 4c(−a+b+c).
4c(−a+b+c)−[3a(a+b+c)−2b(a−b+c)]
=−4ac+4bc+4c2−[3a2+3ab+3ac−2ab+2b2−2bc]
=−4ac+4bc+4c2−[3a2+2b2+3ab−2bc+3ac−2ab]
=−4ac+4bc+4c2−[3a2+2b2+ab+3ac−2bc]
=−4ac+4bc+4c2−3a2−2b2−ab−3ac+2bc
=−3a2−2b2+4c2−ab+4bc+2bc−4ac−3ac
=−3a2−2b2+4c2−ab+6bc−7ac