To express \( x^5 - \frac{1}{x^5} \) as a polynomial in \( x - \frac{1}{x} \), we will first define a variable \( y = x - \frac{1}{x} \).
Using the identity for the powers of \( x - \frac{1}{x} \), we can derive a few key terms to help us find \( p(y) \).
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First, we compute \( x^2 + \frac{1}{x^2} \): \[ \left(x - \frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2} \implies y^2 = x^2 - 2 + \frac{1}{x^2}, \] Thus, \[ x^2 + \frac{1}{x^2} = y^2 + 2. \]
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Next, we compute \( x^3 + \frac{1}{x^3} \): \[ x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) - \left(x + \frac{1}{x}\right). \] We know that \( x + \frac{1}{x} = y + 2 \). Now using \( x^2 + \frac{1}{x^2} = y^2 + 2 \): \[ x^3 + \frac{1}{x^3} = (y + 2)(y^2 + 2) - (y + 2). \] Simplifying this: \[ = (y + 2)(y^2 + 2) - (y + 2) = (y + 2)(y^2 + 2 - 1) = (y + 2)(y^2 + 1) = y^3 + 3y + 2. \]
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We can compute \( x^4 + \frac{1}{x^4} \) now using \( (x^2 + \frac{1}{x^2})^2 - 2 \): \[ x^4 + \frac{1}{x^4} = (y^2 + 2)^2 - 2 = y^4 + 4y^2 + 4 - 2 = y^4 + 4y^2 + 2. \]
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Finally, we compute \( x^5 + \frac{1}{x^5} \): \[ x^5 + \frac{1}{x^5} = (x + \frac{1}{x})(x^4 + \frac{1}{x^4}) - (x^3 + \frac{1}{x^3}) = (y + 2)(y^4 + 4y^2 + 2) - (y^3 + 3y + 2). \] Expanding this, we need to compute \( (y + 2)(y^4 + 4y^2 + 2) \): \[ = y^5 + 4y^3 + 2y + 2y^4 + 8y^2 + 4. \] Therefore, \[ x^5 + \frac{1}{x^5} = y^5 + 4y^3 + 2y + 2y^4 + 8y^2 + 4 - (y^3 + 3y + 2) = y^5 + 3y^3 + 5y + 2. \]
Now, we note that: \[ x^5 - \frac{1}{x^5} = (x^5 + \frac{1}{x^5}) - 2. \] Thus, we have: \[ x^5 - \frac{1}{x^5} = y^5 + 3y^3 + 5y + 2 - 2 = y^5 + 3y^3 + 5y. \]
Therefore, \( p(y) \) such that \( p(x - \frac{1}{x}) = x^5 - \frac{1}{x^5} \) is: \[ p(y) = y^5 + 3y^3 + 5y. \]