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Question

Evaluate (8a3−27b3)(64x3+y3), where (64x3+y3)≠0.


A

(2a3b)(4a2+6ab+9b2)(4x+y)(16x24xy+y2)

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B

(2a3b)(4a2+6ab+9b2)(4xy)(16x24xy+y2)

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C

(2a+3b)(4a2+6ab+9b2)(4x+y)(16x24xy+y2)

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D

(2a+3b)(4a2+6ab+9b2)(4xy)(16x24xy+y2)

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Solution

The correct option is A

(2a3b)(4a2+6ab+9b2)(4x+y)(16x24xy+y2)


Using the identity a3+b3=(a+b)(a2ab+b2)
(64x3+y3)=(4x+y)(16x24xy+y2)
Using Identity a3b3=(ab)(a2+ab+b2)
(8a327b3)=(2a3b)(4a2+6ab+9b2)

(8a327b3)(64x3+y3)=(2a3b)(4a2+6ab+9b2)(4x+y)(16x24xy+y2)


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