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 +b²+c ² +2ab+2bc+2ca=(a+b+c) ²]
=16a ² +b ² +4c ² −8ab−4ac+16ca
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