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Question

Factorise:
81x33y3

A
3(3xy)[9x2+3xy+y2]
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B
3(9xy)[9x2+3xy+y2]
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C
(3xy)[9x2+3xy+y2]
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D
3(3xy)[9x2+6xy+y2]
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Solution

The correct option is A 3(3xy)[9x2+3xy+y2]
81x33y3=3(27x3y3)
=3(33x3y3)
=3[(3x)3y3]
Comparing this with a3b3,
we get a = 3x, b = y
Using the identity,
a3b3=(ab)(a2+ab+b2)
3(3xy)[(3x)2+(3x)(y)+y2]
81x33y3=3(3xy)[9x2+3xy+y2]

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