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Question

Factorise 121b2−88bc+16c2

A
(11b4c)2
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B
(11b44c)2
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C
(b44c)2
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D
(11b+4c)2
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Solution

The correct option is D (11b4c)2
121b288bc+16c2
=(11b)22(11b)(4c)+(4c)2
=(11b4c)2 Using,(ab)2=a2+b22ab

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