9) \( X^{6}-7 X^{3}-8 \)

We start with the polynomial \[ X^{6} - 7X^{3} - 8. \] Notice that the exponent 6 is twice 3. This suggests the substitution \[ y = X^3. \] Then the polynomial becomes \[ y^2 - 7y - 8. \] Next, we factor the quadratic \(y^2 - 7y - 8\). We look for two numbers whose product is \(-8\) and whose sum is \(-7\). These numbers are \(-8\) and \(1\), so we have: \[ y^2 - 7y - 8 = (y - 8)(y + 1). \] Now, substitute back \(y = X^3\) to obtain: \[ (X^3 - 8)(X^3 + 1). \] Both factors are cubic expressions that can be factored further using the formulas for the difference and sum of cubes: - For the difference of cubes: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2), \] with \(a = X\) and \(b = 2\), so \[ X^3 - 8 = (X - 2)(X^2 + 2X + 4). \] - For the sum of cubes: \[ a^3 + b^3 = (a + b)(a^2 - ab + b^2), \] with \(a = X\) and \(b = 1\), so \[ X^3 + 1 = (X + 1)(X^2 - X + 1). \] Thus, the complete factorization of the original polynomial is: \[ X^6 - 7X^3 - 8 = (X - 2)(X^2 + 2X + 4)(X + 1)(X^2 - X + 1). \]