Expand the following:
(i) (4Given that: (4a − b + 2c) 2= (4a) 2 + (−b) 2 + (2c)2 + 2(4a)(−b) + 2(−b)(2c) + 2(2c)(4a)[∵ (a + b + c) 2 = a2 + b2 + c2 + 2ab + 2bc + 2ca)]= 16a2 + b2 + 4c2 – 8ab – 4bc + 16ca(ii). Given that: (3
Given that: (4a − b + 2c) 2
= (4a) 2 + (−b) 2 + (2c)2 + 2(4a)(−b) + 2(−b)(2c) + 2(2c)(4a)
[∵ (a + b + c) 2 = a2 + b2 + c2 + 2ab + 2bc + 2ca)]
= 16a2 + b2 + 4c2 – 8ab – 4bc + 16ca
(ii). Given that: (3