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Factor completely:

8x^(8)-x^(4)a^(4)-9a^(8)
Answer:

Factor completely:\newline8x8x4a49a8 8 x^{8}-x^{4} a^{4}-9 a^{8} \newlineAnswer:

Full solution

Q. Factor completely:\newline8x8x4a49a8 8 x^{8}-x^{4} a^{4}-9 a^{8} \newlineAnswer:
  1. Identify Factoring Technique: We are looking to factor the polynomial 8x8x4a49a88x^{8}-x^{4}a^{4}-9a^{8}. To start, we can try to identify a common pattern or structure in the polynomial that might suggest a particular factoring technique. One technique that might be useful here is factoring by grouping, since we have three terms. We can also look for a common factor in all terms, but there is none in this case. Let's proceed with factoring by grouping.
  2. Grouping Terms: We can group the terms as follows: 8x8x4a48x^{8}-x^{4}a^{4} - 9a89a^{8}. Now we will look for common factors within each group. The first group has a common factor of x4x^{4}, and the second group is a single term which is a perfect square.
  3. Factor Out Common Factor: Factoring out the common factor of x4x^{4} from the first group gives us x4(8x4a4)x^{4}(8x^{4}-a^{4}). The second group is already a perfect square, 9a8-9a^{8}, which can be written as (3a4)2-(3a^{4})^2. Now our expression looks like this: x4(8x4a4)(3a4)2x^{4}(8x^{4}-a^{4}) - (3a^{4})^2.
  4. Apply Difference of Squares: We notice that the expression inside the parentheses, 8x4a48x^{4}-a^{4}, is a difference of squares since 8x48x^{4} is (2x2)2(2x^{2})^{2} and a4a^{4} is (a2)2(a^{2})^{2}. The difference of squares can be factored as (A2B2)=(A+B)(AB)(A^{2} - B^{2}) = (A + B)(A - B). Applying this to 8x4a48x^{4}-a^{4}, we get (2x2)2(a2)2=(2x2+a2)(2x2a2)(2x^{2})^{2} - (a^{2})^{2} = (2x^{2} + a^{2})(2x^{2} - a^{2}).
  5. Correct Error: Now we substitute back into our expression with the factored form of 8x4a48x^{4}-a^{4}. We get x4((2x2+a2)(2x2a2))(3a4)2x^{4}((2x^{2} + a^{2})(2x^{2} - a^{2})) - (3a^{4})^{2}. However, we made a mistake in the previous step; we did not correctly apply the difference of squares to the term 9a8-9a^{8}. The correct factorization of 9a8-9a^{8} should be ((3a4)2)-((3a^{4})^{2}), which is already in the factored form as a difference of squares. We need to correct this error.

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