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Question

Simplify : a216a38×2a23a22a2+9a+4÷3a211a4a2+2a+4

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Solution

We know, a216=a242=(a+4)(a4)
and a38=a323=(a2)(a2+2a+4)
Therefore, a216a318×2a23a22a2+9a+4÷3a211a4a2+2a+4
=a216a316×2a23a22a2+9a+4×a2+2a+43a2+11a4
=(a+4)(a4)(a2)(a2+2a+4)×(2a+1)(a2)(2a+1)(a+4)×a2+2a+4(3a+1)(a4)
=13a+1

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