Find the value of 183-73-3-18-7-2536-6-243-2-15-81-4-20-27-8-15-9-16-6-3-32-64

The numerator is of the form nbsp; a ^ 3 + b ^ 3 nbsp;+ 3ab(a + b) = (a + b)^3 ∴ N' = (18 + 7)^ 3 = 25 ^ 3 ∴ D' = 3^ 6 + ^6 nbsp;C nbsp;_1 · 3^5 · ...

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Byju's Answer

Standard XII

Mathematics

Binomial Theorem

Find the valu...

Question

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

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Solution

The correct option is A
1

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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Similar questions

Q. The value of

\frac{18^{3} + 7^{3} + 3 \cdot 18 \cdot 7 \cdot 25}{3^{6} + 6 \cdot 243 \cdot 2 + 15 \cdot 181 \cdot 4 + 20 \cdot 27 \cdot 8 + 15 \cdot 9 \cdot 16 + 6 \cdot 3 \cdot 32 + 64}

Prepare a frequency distribution by inclusive method taking class interval of 7 from the following data:

28	17	15	22	29	21	23	27	18	12	7	2	9	4
1	8	3	10	5	20	16	12	8	4	33	27	21	15
3	36	27	18	9	2	4	6	32	31	29	18	14	13
15	11	9	7	1	5	37	32	28	26	24	20	19	25
19	20	6	9

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

Q. Find the sum of

1 + \frac{2}{3} + \frac{4}{9} + \frac{6}{27} + \frac{8}{81} + \frac{10}{243} + \dots \infty

The value of $\frac{(18^{3} + 7^{3} + 3.18.7.25)}{3^{6} + 6.243.2 + 15.81.4 + 20.27.8 + 15.9.16 + 6.3.32 + 64}$ is

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28	17	15	22	29	21	23	27	18	12	7	2	9	4
1	8	3	10	5	20	16	12	8	4	33	27	21	15
3	36	27	18	9	2	4	6	32	31	29	18	14	13
15	11	9	7	1	5	37	32	28	26	24	20	19	25
19	20	6	9

28	17	15	22	29	21	23	27	18	12	7	2	9	4
1	8	3	10	5	20	16	12	8	4	33	27	21	15
3	36	27	18	9	2	4	6	32	31	29	18	14	13
15	11	9	7	1	5	37	32	28	26	24	20	19	25
19	20	6	9

Find the value of (183+73+3⋅18⋅7⋅25)36+6⋅243⋅2+15⋅81⋅4+20⋅27⋅8+15⋅9⋅16+6⋅3⋅32+64

The correct option is A 1 The numerator is of the form a3 + b3+3ab(a+b)=(a+b)3 ∴ N' = (18+7)3 = 253 ∴ D' = 36+6C1⋅35⋅21+6C2⋅34⋅22+6C3⋅33⋅23+6C4⋅32⋅24+6C5⋅31⋅25+6C6⋅26 This is clearly the expansion of (3+2)6=56=(25)3 N′D′=(25)3(25)3=1

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

28	17	15	22	29	21	23	27	18	12	7	2	9	4
1	8	3	10	5	20	16	12	8	4	33	27	21	15
3	36	27	18	9	2	4	6	32	31	29	18	14	13
15	11	9	7	1	5	37	32	28	26	24	20	19	25
19	20	6	9