Question

How do you solve `4( x - 3) ^ { 3} + 12= 120`?

Answer 1

Real root `x=6`

Complex roots: `x=3/2+-(3sqrt(3))/2i`

Explanation:

The difference of cubes can be written:

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

We can use this with `a=(x-3)` and `b=3`.

Given:

`4(x-3)^3+12 = 120`

Note that all of the coefficients are divisible by `4`, so first divide both sides by `4` to get:

`(x-3)^3+3 = 30`

Transpose and subtract `30` from both sides to get:

`0 = (x-3)^3-27`

`color(white)(0) = (x-3)^3-3^3`

`color(white)(0) = ((x-3)-3)((x-3)^2+3(x-3)+3^2)`

`color(white)(0) = (x-6)(x^2-6x+9+3x-9+9)`

`color(white)(0) = (x-6)(x^2-3x+9)`

So the real root of our cubic equation is `x=6`

The complex roots are the zeros of `x^2-3x+9`, which is in the standard form `ax^2+bx+c` with `a=1`, `b=-3` and `c=9`. This has zeros given by the quadratic formula:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

`color(white)(x) = (3+-sqrt(3^2-4(1)(9)))/(2*1)`

`color(white)(x) = (3+-sqrt(9-36))/2`

`color(white)(x) = (3+-sqrt(-27))/2`

`color(white)(x) = 3/2+-(3sqrt(3))/2i`

`color(white)()`
Alternative method

Note that the cube roots of `1` are `1`, `omega` and `omega^2 = bar(omega)` where:

`omega = -1/2+sqrt(3)/2i`

So the equation:

`t^3=c^3`

has roots `t = c`, `t = omegac` and `t = omega^2c`.

So given:

`4(x-3)^3+12 = 120`

Subtract `12` from both sides to get:

`4(x-3)^3 = 108`

Divide both sides by `4` to get:

`(x-3)^3 = 27 = 3^3`

which has roots:

`(x-3) = 3`

`(x-3) = 3omega = -3/2+(3sqrt(3))/2i`

`(x-3) = 3omega^2 = -3/2-(3sqrt(3))/2i`

Adding `3` to both sides:

`x = 6`

`x = 3/2+(3sqrt(3))/2i`

`x = 3/2-(3sqrt(3))/2i`