Question

How do you factor `8a^3+27`?

Answer 1

`(2a+3)(4a^2-6a+9)`

Explanation:

We can find one root immediately:
`8a^3+27=0rArr8a^3=-27rArr2a=-3`

So factor out linear bracket `2a+3`:
`8a^3+27=(2a+3)(4a^2-6a+9)`

We know from the fact that this cubic expression has no quadratic or linear terms (and can hence be solved as above) that the other two of the three roots will be complex conjugate values (i.e. `a+-bi`) for points equally spaced around the origin on the complex plane. In other words - they are not real numbers, and so there are no further linear factors.

We can see this also by considering the discriminant of the quadratic factor, `B^2-4AC`, the quantity inside the square root in the quadratic equation. In this case, `B^2-4AC=36-4*4*9=-108`, so the quadratic formula square root produces an imaginary component and there are no real roots to the quadratic bracket.

Answer 2

`(2a+3)(4a^2-6a+9)`

Explanation:

Note that `8a^3` and `27` are both perfect cubes, since `8a^3 = (2a)^3` and `27 = 3^3`.

There is a rule to factor the sum of two cubes:

`x^3+y^3 = (x+y)(x^2-xy+y^2)`

Just for completess, although you don't need it for this exercise, there is a similar formula for the difference of two cubes:

`x^3-y^3 = (x-y)(x^2+xy+y^2)`

In your case, `x = 2a` and `y = 3`.

Plus these values into the formula to get

`(2a)^3+3^3 = (2a+3)((2a)^2-2a*3+(3)^2)`

Rewrite it as

`(2a+3)(4a^2-6a+9)`

Answer 3

THe answer is `=(2a+3)(4a^2-6a+9)`

Explanation:

Apply

`x^3+y^3=(x+y)(x^2-xy+y^2)`

Therefore,

`8a^3+27=(2a)^3+(3)^3`

`=(2a+3)((2a)^2-(2a)(3)+(3)^2)`

`=(2a+3)(4a^2-6a+9)`