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Question

Find the value of x3+27y3+9x2y+27xy2 when x=55 and y=15.


A

772544

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B

1000000

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C

100000000

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D

778342

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Solution

The correct option is B

1000000


The given expression is similiar to the expansion of the identity (a+b)3, i.e.,

x3+27y3+9x2y+27xy2=x3+(3y)3+3×x2×3y+3×x×(3y)2=(x+3y)3

Substituting the values of x and y in (x+3y)3

(x+3y)3=(55+3×15)3=1003=1000000


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