Find the value of (183+73+3⋅18⋅7⋅25)36+6⋅243&#

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Question

Find the value of

\frac{(18^{3} + 7^{3} + 3 \cdot 18 \cdot 7 \cdot 25)}{3^{6} + 6 \cdot 243 \cdot 2 + 15 \cdot 81 \cdot 4 + 20 \cdot 27 \cdot 8 + 15 \cdot 9 \cdot 16 + 6 \cdot 3 \cdot 32 + 64}

Options:

A . 1

B . 5

C . 25

D . 100

Answer: Option A
:
A
The numerator is of the form

a^{3}

+

b^{3} + 3 a b (a + b) = (a + b)^{3}

∴ N' =

(18 + 7)^{3}

=

25^{3}

∴ D' =

3^{6} +^{6} C_{1} \cdot 3^{5} \cdot 2^{1} +^{6} C_{2} \cdot 3^{4} \cdot 2^{2} +^{6} C_{3} \cdot 3^{3} \cdot 2^{3} +^{6} C_{4} \cdot 3^{2} \cdot 2^{4} +^{6} C_{5} \cdot 3^{1} \cdot 2^{5} +^{6} C_{6} \cdot 2^{6}

This is clearly the expansion of

(3 + 2)^{6} = 5^{6} = (25)^{3}

\frac{N^{'}}{D^{'}} = \frac{(25)^{3}}{(25)^{3}} = 1

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