sqdancefan

Answer:

  see below

Step-by-step explanation:

When in doubt, you can always multiply them out to see .

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The expression a^6 + b^6 can be treated as the sum of cubes, which factors as ...

  p^3 + q^3 = (p +q)(p^2 -pq +q^2)

In this case, you have p=a^2 and q=b^2, so the factorization is ...

  a^6 +b^6 = (a^2 +b^2)((a^2)^2 -(a^2)(b^2) +(b^2)^2)

  a^6 +b^6 = (a^2 +b^2)(a^4 -a^2b^2 +b^4) . . . . . . matches choice D

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The expression a^6 -b^6 can be treated either as the difference of cubes, or as the difference of squares. In the former case, its factorization would be ...

  a^6 -b^6 = (a^2 -b^2)(a^4 +a^2b^2 +b^4) . . . . . . matches no offering

In the latter case, its factorization could be any of ...

  a^6 -b^6 = (a^3 -b^3)(a^3 +b^3)

  a^6 -b^6 = (a^3 -b^3)(a +b)(a^2 -ab +b^2) . . . . . . . . . . . matches no offering

  a^6 -b^6 = (a -b)(a^2 +ab +b^2)(a +b)(a^2 -ab +b^2) . . . matches no offering

  a^6 -b^6 = (a^2 -b^2)(a^2 +ab +b^2)(a^2 -ab +b^2) . . . matches no offering