Question

How do you factor `27m^3+1`?

Answer 1

`(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)`

Answer 2

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)`.