Question

hey guys 30+30= 60 points ,is there anyone who can solve/ans this type question with shortest method/any trick..etc?

look once into the attachment provided.
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Answer 1

Multiply and divide by x-1

in numerator

( x-1) ( x+1) = x^2 - 1

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

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

this way .....

x^128 -1 )( x^128 +1) = x^256 -1

So ITS

x^256 -1)/ x-1

Answer 2

Answer :

" Option (4) " is correct.

Solution :

Now,

$(x^{128}+1)(x^{64}+1)(x^{32}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)$

To simplify this form, we multiply (x - 1) to both the numerator and the denominator in order to get a form like $\bigg(\frac{x^{2n}-1}{x-1}\bigg)$, where $n\in \mathbb{Z}$

$\to \frac{(x^{128}+1)(x^{64}+1)(x^{32}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)(x-1)}{(x-1)}$

$=\frac{(x^{128}+1)(x^{64}+1)(x^{32}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)\{(x+1)(x-1)\}}{(x-1)}$

$=\frac{(x^{128}+1)(x^{64}+1)(x^{32}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)\{(x^{2}+1)(x^{2}-1)\}}{(x-1)}$

where $(x+1)(x-1)=(x^{2}-1)$

$=\frac{(x^{128}+1)(x^{64}+1)(x^{32}+1)(x^{16}+1)(x^{8}+1)\{(x^{4}+1)(x^{4}-1)\}}{(x-1)}$

where $(x^{2}+1)(x^{2}-1)=(x^{4}-1)$

$=\frac{(x^{128}+1)(x^{64}+1)(x^{32}+1)(x^{16}+1)\{(x^{8}+1)(x^{8}-1)\}}{(x-1)}$

where $(x^{4}+1)(x^{4}-1)=(x^{8}-1)$

$=\frac{(x^{128}+1)(x^{64}+1)(x^{32}+1)\{(x^{16}+1)(x^{16}-1)\}}{(x-1)}$

where $(x^{8}+1)(x^{8}-1)=(x^{16}-1)$

$=\frac{(x^{128}+1)(x^{64}+1)\{(x^{32}+1)(x^{32}-1)\}}{(x-1)}$

where $(x^{16}+1)(x^{16}-1)=(x^{32}-1)$

$=\frac{(x^{128}+1)\{(x^{64}+1)(x^{64}-1)\}}{(x-1)}$

where $(x^{32}+1)(x^{32}-1)=(x^{64}-1)$

$=\frac{(x^{128}+1)(x^{128}-1)}{(x-1)}$

where $(x^{64}+1)(x^{64}-1)=(x^{128}-1)$

$=\frac{(x^{256}-1)}{(x-1)}$