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(9) \( \quad\left(1-a^{4}\right)\left(1-a^{2}\right)\left(1+a^{2}\right) \)

Ask by Reid Perkins. in South Africa
Feb 03,2025

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Answer

\[ \left(1 - a^{4}\right)\left(1 - a^{2}\right)\left(1 + a^{2}\right) = (1 - a^{2})^{2}(1 + a^{2})^{2} \]

Solution

To simplify the expression: \[ \left(1 - a^{4}\right)\left(1 - a^{2}\right)\left(1 + a^{2}\right) \] we can follow these steps: 1. **Factor \(1 - a^{4}\):** Notice that \(1 - a^{4}\) is a difference of squares: \[ 1 - a^{4} = (1 - a^{2})(1 + a^{2}) \] 2. **Substitute the factorization back into the original expression:** \[ \left(1 - a^{4}\right)\left(1 - a^{2}\right)\left(1 + a^{2}\right) = \left[(1 - a^{2})(1 + a^{2})\right]\left(1 - a^{2}\right)\left(1 + a^{2}\right) \] 3. **Rearrange and combine like terms:** \[ (1 - a^{2}) \times (1 - a^{2}) \times (1 + a^{2}) \times (1 + a^{2}) = (1 - a^{2})^{2} \times (1 + a^{2})^{2} \] 4. **Final Factored Form:** \[ (1 - a^{2})^{2}(1 + a^{2})^{2} \] Alternatively, this can be fully factored further over the real numbers: \[ (1 - a)^2(1 + a)^2(1 + a^{2})^2 \] **Summary:** \[ \left(1 - a^{4}\right)\left(1 - a^{2}\right)\left(1 + a^{2}\right) = (1 - a^{2})^{2}\,(1 + a^{2})^{2} \] Or, more completely factored: \[ (1 - a)^2\, (1 + a)^2\, (1 + a^{2})^2 \]

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Mind Expander

To analyze the expression \( (1-a^4)(1-a^2)(1+a^2) \), we can first simplify it using algebraic identities. Start by breaking it down step by step: 1. Notice that \( (1-a^4) \) can be factored further: \[ 1-a^4 = (1-a^2)(1+a^2) \] So we can rewrite the entire expression as: \[ (1-a^4)(1-a^2)(1+a^2) = ((1-a^2)(1+a^2))(1-a^2)(1+a^2) \] 2. Next, observe that we now have \( (1-a^2)^2(1+a^2)^2 \). So we can simply combine the powers: \[ (1-a^2)^2(1+a^2)^2 = ((1-a^2)(1+a^2))^2 \] 3. The expression \( (1-a^2)(1+a^2) \) simplifies to: \[ 1 - a^4 \] Putting everything together, we simplify it as follows: \[ (1-a^4)^2 \] Thus, the final simplified expression is: \[ \boxed{(1-a^4)^2} \]

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