# Also indices

1. Dec 16, 2009

### 1/2"

1. The problem statement, all variables and given/known data
The problem is
If x= 2+2 2/3 +2 1/3
solve
x 3-6x 2+ 6x-2 =0

2. Relevant equations

3. The attempt at a solution
First i tried to substitute the values but its turning out to be really big( and i get quite scared when it turns up like this and so i didn't go any further this way)
then I tried to make a cube of x-1
Like this
x3 -1 -3x 2+ 3x-3x 2+3x-1+x 3=0
<==>(x-1)3+(x-1)3=x 3
<==>2(x-1)3=x 3
Well I can't get any further!!
Happy if anyone helps!!

Last edited: Dec 16, 2009
2. Dec 16, 2009

### Staff: Mentor

There's something screwy in your problem description. It's akin to saying "If x = 3, how many things are in a dozen?" There doesn't seem to be any connection between the two statements. Are we supposed to assume that x = 2 + 22/3 + 21/3 is a solution of the second equation?

3. Dec 16, 2009

### Staff: Mentor

Why did you pick x - 1? I can't think of any good reason to do this.

Also, I have no idea of what you're trying to do in the work below.

4. Dec 16, 2009

### LCKurtz

I presume you mean "show", not "solve". You are trying to show that value of x is a root of the cubic.

I think you just quit too soon.
$$2 = \left (\frac {x}{x-1}\right)^3$$
$$\frac x {x-1} = 2^{\frac 1 3}$$

Solve for x:

$$x = \frac{2^{\frac 1 3}}{2^{\frac 1 3}-1}$$

Now all you have to do is show this is equal to 2+2 2/3 +2 1/3