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### 3 Answers

- PaulR2Lv 78 years agoFavorite Answer
Let u² = x^12

(u² - 64)

(u + 8)(u - 8)

(x^6 + 8)(x^6 - 8)

Let u³ = x^6

(u³ + 8)(u³ - 8)

(u³ + 8)(u - 2)(u² + 2u - 4)

(u + 2)(u² - 2u + 4)(u - 2)(u² + 2u + 4)

(x² + 2)(x^4 - 2x² + 4)(x² - 2)(x^4 + 2x² + 4)

At this point all our terms are prime and cannot be factored anymore. The proof of this is that after factoring cubic, the trinomial is always prime. You should also take notice that the binomials are prime because their constants are prime. I hope this helps. Have a good day.

- PolyhymnioLv 78 years ago
Factor as the difference of two squares

(x^6 + 8)(x^6 - 8).

Now factor as the sum and difference of two cubes

(x^2 + 2)(x^4 - 2x^2 + 4)(x^2 - 2)(x^4 + 2x^2 + 4)

- ranjankarLv 78 years ago
x^12 - 64

USE FORMULA a^2 - b^2 = ( a+b) (a-b)

( x^6 + 8) ( x^6 - 8 )

USE FORMULAS a^3 + b^3 = ( a+b) ( a^2 -ab + b^2 )

AND a^3 - b^3 = ( a-b) ( a^2 + ab + b^2)

( x^2 + 2) ( x^4 - 2x^2 +4) ( x^2 - 2) ( x^4 + 2x^2 + 4 ) ANSWER