can m^3 - 8 be factored?

Answer 1

Hey there!

Yes! Even though it may not seem it, you can apply a formula.

This formula is the Difference of Two Cubes, which states:

m^3-n^3=(m-n)(m^2+mn+n^2).

So.

m^3-2^3=(m-2)(m^2+2m+2^2) -->
(m-2)(m^2+2m+4)

So the answer is (m-2)(m^2+2m+4).

Hope it helps!

Answer 2

Yes. One law of factoring is that

(x – y)(x^2 + xy + y^2) = x^3 + (x^2 * y) + (x * y^2) – (x^2 * y) – (x ^ y^2) – y^3

for any x and y. The middle four terms on the right side ("right side" means everything to the right of the equal sign =) all cancel out, so we can simplify that to

x^3 – y^3 = (x – y)(x^2 + xy + y^2).

Since you have m^3 – 8 and 2^3 = 8, substitute x and y with m and 2 respectively. And so

m^3 – 8 = (m – 2)(m^2 + 2m + 4).

Answer 3

(m^3-8)= (m-2)(m^2+2m+4)so the -4m and 4m cancel out and the 2m^2 and 2m^2 cancel out leaving you with m^3-8

Answer 4

yes

(m - 2)(m^2 + 2m + 4)

Answer 5

yes...

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

(m-2)(m^2 +2m + 4)

Answer 6

I believe it cannot be factored into anything.