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Find HCF and LCM in Simplified form using Factorization method. \[ x^{6}-1, x^{4}+x^{2}+1 \]

Ask by Vega Rodriquez. in the United Arab Emirates
Mar 20,2025

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The HCF of \( x^{6}-1 \) and \( x^{4}+x^{2}+1 \) is 1, and the LCM is \( (x^{6}-1)(x^{4}+x^{2}+1) \).

Solution

To find the Highest Common Factor (HCF) and the Least Common Multiple (LCM) of the polynomials \( x^{6}-1 \) and \( x^{4}+x^{2}+1 \) using the factorization method, we will follow these steps: ### Step 1: Factorize \( x^{6}-1 \) The expression \( x^{6}-1 \) can be factored using the difference of squares and the difference of cubes: \[ x^{6}-1 = (x^{3}-1)(x^{3}+1) \] Next, we can factor \( x^{3}-1 \) and \( x^{3}+1 \): \[ x^{3}-1 = (x-1)(x^{2}+x+1) \] \[ x^{3}+1 = (x+1)(x^{2}-x+1) \] Thus, the complete factorization of \( x^{6}-1 \) is: \[ x^{6}-1 = (x-1)(x^{2}+x+1)(x+1)(x^{2}-x+1) \] ### Step 2: Factorize \( x^{4}+x^{2}+1 \) The expression \( x^{4}+x^{2}+1 \) can be treated as a quadratic in terms of \( x^{2} \): Let \( y = x^{2} \), then we have: \[ y^{2} + y + 1 \] To factor \( y^{2} + y + 1 \), we can use the quadratic formula: \[ y = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \] This means \( y^{2} + y + 1 \) does not factor over the reals, but we can express it in terms of its roots: \[ x^{4}+x^{2}+1 = (x^{2} - \frac{-1 + i\sqrt{3}}{2})(x^{2} - \frac{-1 - i\sqrt{3}}{2}) \] However, for our purposes, we can keep it as \( x^{4}+x^{2}+1 \) since it does not share any common factors with \( x^{6}-1 \). ### Step 3: Find HCF and LCM #### HCF The HCF is the product of the common factors. Since \( x^{4}+x^{2}+1 \) does not share any factors with \( x^{6}-1 \), we have: \[ \text{HCF}(x^{6}-1, x^{4}+x^{2}+1) = 1 \] #### LCM The LCM is given by the product of the two polynomials divided by their HCF: \[ \text{LCM}(x^{6}-1, x^{4}+x^{2}+1) = \frac{(x^{6}-1)(x^{4}+x^{2}+1)}{\text{HCF}(x^{6}-1, x^{4}+x^{2}+1)} \] Since the HCF is 1, we have: \[ \text{LCM}(x^{6}-1, x^{4}+x^{2}+1) = (x^{6}-1)(x^{4}+x^{2}+1) \] ### Final Results - **HCF**: \( 1 \) - **LCM**: \( (x^{6}-1)(x^{4}+x^{2}+1) \) This concludes the factorization and calculation of HCF and LCM for the given polynomials.

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To find the HCF and LCM using the factorization method, we first need to factor the given polynomials. 1. For \( x^6 - 1 \), we can factor it as follows: \[ x^6 - 1 = (x^3 - 1)(x^3 + 1) \] Further factoring gives us: \[ x^3 - 1 = (x - 1)(x^2 + x + 1) \quad \text{and} \quad x^3 + 1 = (x + 1)(x^2 - x + 1) \] Therefore, \[ x^6 - 1 = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1) \] 2. For \( x^4 + x^2 + 1 \), we can substitute \( y = x^2 \): \[ y^2 + y + 1 = \frac{y^3 - 1}{y-1} = \frac{(y - \omega)(y - \omega^2)}{y - 1} \quad \text{(where } \omega = e^{2\pi i/3}\text{)} \] Thus, this does not factor nicely over the reals but can be written in terms of complex roots for a complete factorization. The HCF (Highest Common Factor) would be the product of common factors. Notably, \( x^2 + x + 1 \) appears in both factorizations. The LCM can be calculated by taking the product of all unique factors, which yields: \[ LCM = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1)(x^4 + x^2 + 1) \] Hence, in simplified terms: - HCF: \( x^2 + x + 1 \) - LCM: \( (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1)(x^4 + x^2 + 1) \)

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