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Question

3a6×b3×c21c9×a12=

A
bc3a2
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B
bc4a2
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C
ab4c2
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D
bc4a2
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Solution

3a6×b3×c21c9×a12

=a612×b3×c219 [Since, xmxn=xmn,x0]

=3a6×b3×c12

=3b3×c12a6 [Since,am=1am]

=(b3×c12a6)13

=b3×13×c12×13a6×13 [Since, (am)n=am×n]

=bc4a2

3a6×b3×c21c9×a12=bc4a2

Hence, Option D is correct.


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