How to find roots of (u-3)^3-3u+8?

Question

Here's what I've done so far:
(u-3)^3-3u+8=0
(u-3)(u2-3u+9)-3u+8=0
u3-3u2+9u+92-3u2=27=3u+8=o
u3-9u2+18u-19=0...

but I don't know how to continue on now...

Thanks all in advance =D

Answer 1

First, let's substitute a variable for the value (u-3).

Let w= u-3.

Now we can rewrite the equation this way:

(u-3)^3 -3u +8

= (u-3)^3 -3u +9 -9 +8

= (u-3)^3 -3(u-3) -9 + 8

= (u-3)^3 - 3(u-3) -1

= w^3 -3w -1;

or, w^3 -3w=1

When a cubic equation has zero for the coefficient of the w^2 term, like this equation, it is called a 'depressed cubic' equation. The method for solving was discovered by Niccolo Tartaglia in the 1500s.

you can read more at the Wikipedia link.