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How do you factor #x^6 - 8y^3#?

Answer 1

Eva Adamson

#x^6-8y^3=(x^2-2y)(x^4+2x^2y+4y^2)#

You will find that #a^3-b^3 = (a-b)(a^2+ab+b^2)#

Therefore by setting #a=x^2# and #b=2y#:

#x^6-8y^3=(x^2-2y)(x^4+2x^2y+4y^2)#

Don't think it does but it possibly could be factorised more.

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Answer 2

Genesis Allen

To factor (x^6 - 8y^3), recognize it as a difference of cubes since (x^6 = (x^2)^3) and (8y^3 = (2y)^3). The formula for factoring a difference of cubes (a^3 - b^3) is ((a - b)(a^2 + ab + b^2)). Applying this to (x^6 - 8y^3), let (a = x^2) and (b = 2y), so:

[ x^6 - 8y^3 = (x^2)^3 - (2y)^3 = (x^2 - 2y)((x^2)^2 + (x^2)(2y) + (2y)^2) = (x^2 - 2y)(x^4 + 2x^2y + 4y^2). ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.