Polynomials!!

We are given the equation $p(x - \frac{1}{x}) = x^5 - \frac{1}{x^5}$, where $p(z)$ is a polynomial in terms of $z = x - \frac{1}{x}$. Our goal is to express $x^5 - \frac{1}{x^5}$ as a polynomial in $z$.

### **Solution By Steps**

*Step 1: Express $x^5 - \frac{1}{x^5}$ in terms of powers of $x - \frac{1}{x}$*

We start by recalling some trigonometric identities and symmetries for powers of $x$ and their reciprocals.

Let $z = x - \frac{1}{x}$. We first compute the first few powers of $z$:

$$z = x - \frac{1}{x}$$

Square both sides to compute $z^2$:

$$z^2 = \left(x - \frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2} = x^2 + \frac{1}{x^2} - 2$$

Thus,

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

Next, let's compute $x^3 - \frac{1}{x^3}$:

$$x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) = z \cdot (z^2 + 2) = z^3 + 2z$$

Now, let's compute $x^5 - \frac{1}{x^5}$ by multiplying $x^2 + \frac{1}{x^2}$ and $x^3 - \frac{1}{x^3}$:

$$x^5 - \frac{1}{x^5} = \left(x^3 - \frac{1}{x^3}\right)\left(x^2 + \frac{1}{x^2}\right) = (z^3 + 2z)(z^2 + 2)$$

$$= z^5 + 2z^3 + 2z^3 + 4z = z^5 + 4z^3 + 4z$$

Thus, we have expressed $x^5 - \frac{1}{x^5}$ as:

$$x^5 - \frac{1}{x^5} = z^5 + 4z^3 + 4z$$

*Step 2: Write the polynomial $p(z)$*

From the above expression, we can see that the polynomial $p(z)$ that satisfies $p(x - \frac{1}{x}) = x^5 - \frac{1}{x^5}$ is:

$$p(z) = z^5 + 4z^3 + 4z$$

### **Final Answer**

The polynomial $p(z)$ is:

$$p(z) = z^5 + 4z^3 + 4z$$