What is the simplified value of (x^{128}+1)(x | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) To simplify the expression $(x^{128}+1)(x^{64}+1)(x^{32}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)$,we use the algebraic identity $(a-b)(a+b) =

Vedclass · Accepted Answer

$\frac{x^{256}-1}{x-1}$

Vedclass · Answer

(D) To simplify the expression $(x^{128}+1)(x^{64}+1)(x^{32}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)$,we use the algebraic identity $(a-b)(a+b) = a^2-b^2$.Multiply the entire expression by $\frac{x-1}{x-1}$:$= \frac{(x-1)(x+1)(x^2+1)(x^4+1)(x^8+1)(x^{16}+1)(x^{32}+1)(x^{64}+1)(x^{128}+1)}{x-1}$Using $(x-1)(x+1) = x^2-1$,we get:$= \frac{(x^2-1)(x^2+1)(x^4+1)(x^8+1)(x^{16}+1)(x^{32}+1)(x^{64}+1)(x^{128}+1)}{x-1}$$= \frac{(x^4-1)(x^4+1)(x^8+1)(x^{16}+1)(x^{32}+1)(x^{64}+1)(x^{128}+1)}{x-1}$Continuing this process repeatedly:$= \frac{(x^8-1)(x^8+1)(x^{16}+1)(x^{32}+1)(x^{64}+1)(x^{128}+1)}{x-1}$$= \frac{(x^{16}-1)(x^{16}+1)(x^{32}+1)(x^{64}+1)(x^{128}+1)}{x-1}$$= \frac{(x^{32}-1)(x^{32}+1)(x^{64}+1)(x^{128}+1)}{x-1}$$= \frac{(x^{64}-1)(x^{64}+1)(x^{128}+1)}{x-1}$$= \frac{(x^{128}-1)(x^{128}+1)}{x-1}$$= \frac{x^{256}-1}{x-1}$.

What is the simplified value of $(x^{128}+1)(x^{64}+1)(x^{32}+1)(x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)$?

Similar Questions

$0.002 \times 0.5 = ?$

If $x^{2}+3x+1=0,$ then the value of $x^{3}+\frac{1}{x^{3}}$ is

Find the value of $\frac{(0.064-0.008)(0.16-0.04)}{(0.16+0.08+0.04)(0.4+0.2)^{3}}$.

$1 \frac{3}{4} + 5 \frac{1}{3} + 3 \frac{2}{5} = ?$

The simplification of $\frac{0.67 \times 0.67 \times 0.67 - 0.001}{0.67 \times 0.67 + 0.067 + 0.01}$ gives

Vedclass Test Series

Exam Paper Generator

Online Exam Module