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 · ...

341

You visited us 341 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Binomial Theorem

Find the valu...

Question

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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

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

Suggest Corrections

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

Join BYJU'S Learning Program

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