Find the value of ( 18 3 + 7 3 + 3 ⋅ 18 ⋅ 7 ⋅ 25 ) 3 6 + 6 ⋅ 243 ⋅ 2 + 15 ⋅ 81 ⋅ 4 + 20 ⋅ 27 ⋅ 8 + 15 ⋅ 9 ⋅ 16 + 6 ⋅ 3 ⋅ 32 + 64 Options: A . &nbsp 1 B . &nbsp 5 C . &nbsp 25 D . &nbsp 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 ⋅ 3 5 ⋅ 2 1 + 6 C 2 ⋅ 3 4 ⋅ 2 2 + 6 C 3 ⋅ 3 3 ⋅ 2 3 + 6 C 4 ⋅ 3 2 ⋅ 2 4 + 6 C 5 ⋅ 3 1 ⋅ 2 5 + 6 C 6 ⋅ 2 6 This is clearly the expansion of ( 3 + 2 ) 6 = 5 6 = ( 25 ) 3 N ′ D ′ = ( 25 ) 3 ( 25 ) 3 = 1

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

Discussion Forum : Binomial Theorem

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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