How do you factor 27m^3+1?

2 Answers
Jul 3, 2016

(3m+1)(9m^2-3m+1)

Explanation:

This is a color(blue)"sum of cubes" and is factorised in general as follows.

color(red)(|bar(ul(color(white)(a/a)color(black)(a^3+b^3=(a+b)(a^2-ab+b^2))color(white)(a/a)|)))

27m^3=(3m)^3" and " 1=(1)^3rArra=3m" and " b=1

Substituting these values gives required factors.

rArr27m^3+1=(3m+1)(9m^2-3m+1)

Jul 3, 2016

Using the formula for the sum of cubes.

Explanation:

We can use the formula for the sum of cubes

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

here our cubes are

(3m)^3+1^3=(3m+1)((3m)^2-3m+1)

=(3m+1)(9m^2-3m+1).

The roots of 9m^2-3m+1 are

m=(3\pm sqrt(9-36))/(18)=(3\pm sqrt(-27))/18

=(3\pm i3sqrt(3))/18=(1\pm isqrt(3))/6

then are both complex. This means that the quadratic part cannot be factorized more with real numbers. We can write the irreducible factorization in complex numbers as

(3m+1)(m-(1+isqrt(3))/6)(m-(1-isqrt(3))/6).