CUET PG 2024 — Mathematics PYQ

Given $x^2+\frac{1}{x^2}=2$

$(x^2+\frac{1}{x^2}{)}^2=2^2$

$x^4+2x^2\frac{1}{x^2}+\frac{1}{x^4}=4$

$x^4+2+\frac{1}{x^4}=4$

$x^4+\frac{1}{x^4}=2$

Now $x^8+\frac{1}{x^8}={\left(x^4+\frac{1}{x^4}\right)}^2=2^2$

$x^8+2x^4+\frac{1}{x^8}=4$

$x^8+2+\frac{1}{x^8}=4$

$x^8+\frac{1}{x^8}=2$

$x^{16}+\frac{1}{x^{16}}=2$

$x^{128}+\frac{1}{x^{128}}=2$

$\frac{1}{x^{256}}+\frac{1}{x^{256}}=2$