Expand: (4a – b + 2c)∧2.
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We have,
(4a−b+2c)
2=(4a) 2 +(−b) 2+(2c) 2 +2(4a)(−b)+2(−b)(2c)+2(2c (4a)
[∵a 2 +b 2 +c 2+2ab+2bc+2ca=(a+b+c) 2 ]
=16a 2 +b 2+4c 2 −8ab−4ac+16ca
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