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

I can't follow what you did on the third line. You have too many equal signs and I don't see how you would get 92 from anything.

(u-3)^3 - 3u + 8 = 0
(look up difference of cubes formula if you don't know it.)
u^3 - 9u^2 +27u - 27 - 3u + 8 = 0
u^3 - 9u^2 + 24u - 19 = 0

At this point, you're pretty much stuck. Using P&Q factoring, the only possible rational factors are +/-1 and +/-19, but none of those four answers work. Using the rule of sign changes, we know it has only one root, so it must be irrational. If you're taking calculus you can use Newton's method to find the root. If you're taking any other class, all you can say is "no rational roots" and move on to the next problem.

Answer 2

You can write (u - 3)³ - 3(u-3) - 1

if x = u - 3, you have to resolve x³ - 3x - 1 = 0

Now, it's difficult !!
It's like x³ + px + q = 0 with p = -3 and q = -1

4p³ + 27q² = -81 < 0 then 3 roots

let t = 1/3 arccos [3q*sqrt(3) / (2p *sqrt(-p))]

t = 1/3 arccos (-3sqrt3 / -6sqrt3) = 1/3 arccos(1/2)

t = 1/3 pi/3 = pi/9

now the first root is
x_1 = sqrt(-4p/3) cost = 2 cos(pi/9)

the second is
x_2 = sqrt(-4p/3) cos( t+ 2pi/3) = 2cos(7pi/9)

the third is
x_3 = sqrt(-4p/3) cos( t+ 4pi/3) = 2cos(13pi/9)

and now
u_1 = x_1 + 3, u_2 = x_2 + 3 and u_3 = x_3 + 3