algebra

[giomar]

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

(a)2^16-1
(b)2^16+1
(c)2^32-1
(d)2^128-1
(e)2^16(2^16-1)

[fighter]

Hi the answer to this question should be Option A.
I feel there is a typo error. the 3rd bracket need to be (2^4 + 1)

[giomar]

Hi the answer to this question should be Option A.
I feel there is a typo error. the 3rd bracket need to be (2^4 + 1)

yes it should be like that.Official Answer is A.
please fighter explain how you solve the problem.

[jayashri_rane]

[(2^2-1) (2^2+1)](2^4+1) (2^8+1)
= [(2^4-1) (2^4+1)](2^8+1) //(a+b)(a-b) =a^2-b^2
= (2^8-1)(2^8+1)
=(2^16-1)

[Salila]

giomar ,

substitute 2^4 with x

then you have (2^2-1) (2^2+1) (2^4+1) (2^8+1) = {(2^2)^2 - 1^2} {(x+1)(x^2+1)}
(2^4 -1) (x+1)(x^2+1) = (x-1)(x+1)(x^2+1) =(x^2 - 1)(x^2+1) = x^4 -1

change x back to the original 2^4 to get (2^4)^4 -1 =2^16 -1

hope it helps

regards
salila.

[giomar]

thanks,I get it now