Question

How do you solve `3x^3 - 2[ x - ( 3 - 2x ) ] = 3 ( x - 2 )^2`?

Answer 1

Real solution: `x=1`

Complex solutions: `x in {1, sqrt(2)i, -sqrt(2)i}`

Explanation:

L.S. `=3x^3-2[x-(3-2x)]`

`color(white)("XXX")=3x^3-2[-3+3x]`

`color(white)("XXX")=3x^3-6x+6`

R.S. `=3(x-2)^2`

`color(white)("XXX")=3(x^2-4x+4)`

`color(white)("XXX")=3x^2-12x+12`

Since L.S. `=` R.S.
`3x^3+6x+6=3x^2-12x+12`

`rarr x^3+2x+2=x^2-4x+4`

`rarr color(green)(1)x^3-x^2+2x-color(blue)(2)=0`

In the hope of finding rational roots we try the factors of `color(blue)(2)/color(green)(1)`

and we find `color(red)(x=1)` is a root
`rarr (x-1)` is a factor.

`(x^3-x^2+2x-2) div (x-1) = x^2+2`

`(x^2+2)` has no Real roots (i.e. no Real solutions for `x`)

However, if we allow Complex solutions:
`color(white)("XXX")x=+-sqrt(2)i`