Question

Factorize a³-64b³-27c³-36abc

Answer 1

(i) a3−64b3−27c3−36abc


=
(a)3+(−4b)3+(−3c)3−3×(a)×(−4b)×(−3c)
=(a−4b−3c)[(a)2+(−4b)2+(−3c)2−a(−4b)−(−4b)(−3c)−(−3c)(a)]
[usingtheidentity,a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)]
=(a− 4b−3c)(a2+16b2+9c2+4ab−12bc+3ac)