We are given that $x - \frac{1}{x} = m$. a) We need to find the value of $x^2 + \frac{1}{x^2}$. b) We need to show that $\frac{x^8 + 1}{x^4} = m^4 + 4m^2 + 2$. c) If $x^4 + \frac{1}{x^4} = 119$, we need to prove that $m = \pm 3$.

1. Problem Description

We are given that $x - \frac{1}{x} = m$.

a) We need to find the value of $x^2 + \frac{1}{x^2}$.

b) We need to show that $\frac{x^8 + 1}{x^4} = m^4 + 4m^2 + 2$.

c) If $x^4 + \frac{1}{x^4} = 119$, we need to prove that $m = \pm 3$.

2. Solution Steps

a) We know that $x - \frac{1}{x} = m$. Squaring both sides, we get:

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

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

$x^2 - 2 + \frac{1}{x^2} = m^2$

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

b) We need to find the value of $\frac{x^8 + 1}{x^4}$, which can be written as $x^4 + \frac{1}{x^4}$.

From part (a), we have $x^2 + \frac{1}{x^2} = m^2 + 2$. Squaring both sides, we get:

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

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

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

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

Thus, $\frac{x^8 + 1}{x^4} = x^4 + \frac{1}{x^4} = m^4 + 4m^2 + 2$.

c) We are given that $x^4 + \frac{1}{x^4} = 119$. From part (b), we know that $x^4 + \frac{1}{x^4} = m^4 + 4m^2 + 2$.

So, $m^4 + 4m^2 + 2 = 119$

$m^4 + 4m^2 - 117 = 0$

Let $y = m^2$. Then, $y^2 + 4y - 117 = 0$.

$y^2 + 13y - 9y - 117 = 0$

$y(y + 13) - 9(y + 13) = 0$

$(y - 9)(y + 13) = 0$

So, $y = 9$ or $y = -13$. Since $y = m^2$, and $m$ is real, we must have $y \ge 0$. Therefore, $y = 9$.

$m^2 = 9$

$m = \pm \sqrt{9}$

$m = \pm 3$

3. Final Answer

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

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

c) $m = \pm 3$