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If x#-1, x^16/{(1+x)*(1+x^2)*(1+x^4)*(1+x^8)

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Updated on: 17 Jan 2019, 00:06
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79% (01:31) correct 21% (02:06) wrong based on 287 sessions

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If $$x\neq{-1}$$ , $$\frac{1- x^{16}}{(1+x)*(1+x^2)*(1+x^4)*(1+x^8)}$$ is equivalent to

A. -1
B. 1
C. x
D. 1-x
E. x-1

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Originally posted by NoHalfMeasures on 06 Jun 2014, 03:03.
Last edited by Bunuel on 17 Jan 2019, 00:06, edited 5 times in total.
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06 Jun 2014, 03:43
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3
Bunuel, Thanks for the correction

Numerator

$$1 - x^{16}$$

$$= (1 + x^8) (1 - x^8)$$

$$= (1 + x^8) (1 + x^4) (1 - x^4)$$

$$= (1 + x^8) (1 + x^4) (1 + x^2) (1 - x^2)$$

$$= (1 + x^8) (1 + x^4) (1 + x^2) (1 + x) (1 - x)$$

Denominator

$$(1 + x^8) (1 + x^4) (1 + x^2) (1 + x)$$

Expanded / simplified equation would be

$$\frac{(1 + x^8) (1 + x^4) (1 + x^2) (1 + x) (1 - x)}{(1 + x^8) (1 + x^4) (1 + x^2) (1 + x)}$$

= 1 - x

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06 Jun 2014, 03:28
1
Is this the value we require to find? Kindly confirm

$$1 - \frac{x^{16}}{(1+x)(1+x^2)(1+x^4)(1+x^8)}$$
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06 Jun 2014, 03:31
2
PareshGmat wrote:
Is this the value we require to find? Kindly confirm

$$1 - \frac{x^{16}}{(1+x)(1+x^2)(1+x^4)(1+x^8)}$$

Edited the original post.
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17 Jan 2019, 00:03
NoHalfMeasures wrote:
If $$x\neq{-1}$$ , $$\frac{1- x^{16}}{{(1+x)*(1+x^2)*(1+x^4)*(1+x^8)}$$ is equivalent to

A. -1
B. 1
C. x
D. 1-x
E. x-1

how to pick number in this question? Bunuel
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17 Jan 2019, 00:12
1
rashedBhai wrote:
NoHalfMeasures wrote:
If $$x\neq{-1}$$ , $$\frac{1- x^{16}}{(1+x)*(1+x^2)*(1+x^4)*(1+x^8)}$$ is equivalent to

A. -1
B. 1
C. x
D. 1-x
E. x-1

how to pick number in this question? Bunuel

Pick x = 1. In this case the numerator is 0, thus the whole fraction is 0. Now, plug x = 1 into the options to check which of them give 0. Only D and E fit.

Now, pick x = 2. In this case the numerator is negative and the denominator is positive, thus the whole fraction is negative. Out of D and E, only D is negative if x = 2.

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17 Jan 2019, 00:25
NoHalfMeasures wrote:
If $$x\neq{-1}$$ , $$\frac{1- x^{16}}{(1+x)*(1+x^2)*(1+x^4)*(1+x^8)}$$ is equivalent to

A. -1
B. 1
C. x
D. 1-x
E. x-1

A question in which basic formula works a^2 - b^2 = (a-b) (a+b)

Now you can expand the numerator as

1-x^16 = (1-x) (1+x) (1+x^2) (1+x^4) (1 + x^8)

Which will result in Answer D
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Re: If x#-1, x^16/{(1+x)*(1+x^2)*(1+x^4)*(1+x^8)   [#permalink] 17 Jan 2019, 00:25
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