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NoHalfMeasures
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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NoHalfMeasures
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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Difference of squares:

­
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Now that this thread has mathematical explanations already, let's try to be creative

Question hasn't limited our imagination to choose value of x so let's choose two values (x=0 and 0.1) and substitute

\(x\neq{-1}\), then \(\frac{1- x^{16}}{(1+x)(1+x^2)(1+x^4)(1+x^8)}\)

@x = 0 \(x\neq{-1}\), then \(\frac{1- x^{16}}{(1+x)(1+x^2)(1+x^4)(1+x^8)} = 1\)

i.e. Option B and E REMAIN and others are OUT

@x = 0.1 \(x\neq{-1}\), then \(\frac{1- x^{16}}{(1+x)(1+x^2)(1+x^4)(1+x^8)} < 1\)
Only option D remains hence

Answer: Option D

remotely Related Video:

NoHalfMeasures
If \(x\neq{-1}\), then \(\frac{1- x^{16}}{(1+x)(1+x^2)(1+x^4)(1+x^8)}\) is equal to

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

PS07007
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