Factor the differences of two squares. Be sure to factor completely.

X^{6}-64

Factor the differences of two squares. Be sure to factor completely.

X^{6}-64

Tutors, please sign in to answer this question.

There are two ways at least you can go to factor X^{6}-64.

Method 1. Treat X^{6}-64 as the difference of two squares first.

X^{6}-64

= (x^{3})^{2} - 8^{2}

= (x^{3} + 8)(x^{3} - 8)

= (x+2)(x^{2}-2x+4)(x-2)(x^{2}+2x+4)

Method 2. Treat X^{6}-64 as the difference of two cubes first.

x^{6}-64

= (x^{2})^{3} - (4)^{3}

= (x^{2} - 4)(x^{4} + 4x^{2} + 16)

= (x+2)(x-2)[(x^{2}+4)^{2} - (2x)^{2}], by completing the square

= (x+2)(x-2)(x^{2}+2x+4)(x^{2}-2x+4)

x^{6}-64= (x^{3})^{2 }- (8)^{2
}

= (x^{3}+8) (x^{3}-8)

=(x+2)(x^{2 }-2x+4) (x-2)(x^{2} +2x+4)

Kayla, for any polynomial, [f(x)]^2 - [g(x)]^2, will factor as [f(x)-g(x)] times [f(x)+g(x)], so if you set f(x) = x^3 and g(x) = 8, you can rewrite the equation as [x^3]^2 - [8]^2, which you can factor as the difference of two squares. The factor "completely" part of the question is important because once you get your two terms, you will find that they are both a sum and difference of cubes, which also have a special factoring process. You can find a good illustration of that process (assuming your text does not provide illustrations) at http://www.cliffsnotes.com/study_guide/Sum-or-Difference-of-Cubes.topicArticleId-257309,articleId-257149.html

I'll leave the rest to you, but feel free to email or ask any follow-up questions if this does not make sense. John

Kathleen D.

Middle and high school math tutor

Franklin Square, NY

5.0
(9 ratings)

PJ J.

Ninety-nineth percentile scoring tutor ready to teach SAT and GMAT

New York, NY

5.0
(2 ratings)

Jeffrey G.

Former Med-Student turned Professional Science Tutor

Brooklyn, NY

5.0
(364 ratings)