Factorise :
a²−b²−4ac+4c²
Given (a2−4ac+4c2)−b2=(a2−2a(2c)+(2c)2)−b2=(a−2c)2−b2 [Since, (a−b)2=a2−2ab+b2]=(a−2c+b)(a−2c−b) [Since, a2−b2=(a−b)(a+b)]
Question 92 (xxvi)
Factorise the following using the identity a2−b2=(a+b)(a−b).
(a−b)2−(b−c)2