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Question

4a2(4b2+4bc+c2)


  1. (2a2bc)(2a+2b+c)

  2. (2a2bc)(2a+2bc)

  3. (2a2bc)(2a2b+c)

  4. (2a+2bc)(2a+2b+c)


Solution

The correct option is A

(2a2bc)(2a+2b+c)


Given that 

 4a2(4b2+4bc+c2)
 =  (2a)2((2b+c)2
 =(2a(2b+c))(2a+2b+c)     (  using difference of squares )
 = (2a2bc)(2a+2b+c)

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