How do you factor `z^3-9x^2+27z-27`?

In the general case we have:

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

In our case, we notice that `z^3` and `-27 = (-3)^3` are both perfect cubes.

So try `a=z` and `b = -3` to find:

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

`=z^3-9z^2+27z-27`

Let `f(z) = z^3-9z^2+27z-27`

We can set `f(z)=0` and use the rational roots theorem to find the factors.

If `z` is a root, then `z` must be a factor of `(-27)/1 = -27`.

∴ Possible values of `z` are `±1, ±3, ±9, ±27`.

We can test a possible root by plugging in the value and seeing if `f(z)` becomes zero.

We find that `3` is a root because

`f(3) = 3^3-9×3^2+27×3-27 = 27-81+81-27 = 0`.

If `3` is a root, then `z-3` is a factor of `f(z)`.

We can use synthetic division to divide `f(z)` by `z-3`.

`3|1color(white)(l)-9color(white)(Xll)27color(white)(l)-27`
`color(white)(1)|color(white)(XXl)3color(white)(l)-18color(white)(Xll)27`
`" "stackrel("—————————————)`
`color(white)(Xll)1color(white)(l)-6color(white)(XXl)9color(white)(XXl)color(red)(0)`

We get `z^2-6z+9` with a remainder of zero.

So `f(z) = (z-3)( z^2-6z+9)`.

We can factor the quadratic to get

`z^2-6z+9 = (z-3)(z-3)`.

So `f(z) = (z-3) (z-3)(z-3)`

∴ `z^3-9z^2+27z-27= (z-3)(z-3)(z-3)`