# No idea -- algebraic manipulation involving powers

1. May 31, 2016

### Chaos_Enlightened

1. The problem statement, all variables and given/known data
If x=2+2^1/2+2^2/3. Then x^3-6x^2+6x=?

2. Relevant equations
(A+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca+)

3. The attempt at a solution
x^2= 6+3(2^2/3)+2^4/3+4(2^1/3)

2. May 31, 2016

### Chaos_Enlightened

Should I go on and find x^3

3. May 31, 2016

### Sahil Kukreja

Hint :- take '2' to the left side and then cube both sides.

Last edited: May 31, 2016
4. May 31, 2016

### Chaos_Enlightened

Ok

5. May 31, 2016

### Chaos_Enlightened

Got x^3-6x^2+12x=14+6(2^1/3+2^2/3)

6. May 31, 2016

### Sahil Kukreja

I think there is a calculation mistake here, sorry if i am wrong.

Last edited: May 31, 2016
7. May 31, 2016

### Nidum

Work out the numerical value of x from the first equation .

What do you do then ?

Last edited: May 31, 2016
8. May 31, 2016

### Sahil Kukreja

The question is solvable without calculating square roots and cube roots.

9. May 31, 2016

### Delta²

factorize the polynomial, if all go well (i.e the polynomial has 3 real roots) you should get it in the form $(x-r_1)(x-r_2)(x-r_3)$

You can easily see that one root of the polynomial is....

Last edited: May 31, 2016
10. May 31, 2016

### Nidum

Think about what sort of answer the question really requires .

11. May 31, 2016

### Chaos_Enlightened

That might just do it

12. May 31, 2016

### Chaos_Enlightened

Which one should I factorise

13. May 31, 2016

### Delta²

$x^3-6x^2+6x=x(x^2-...+...)$ the polynomial inside the parenthesis is a 2nd order polynomial i believe you know how to factorize it,

14. May 31, 2016

### Chaos_Enlightened

Wait
I could just substitute then
X^3-6x^2+12x=14+6(2^1/3+2^2/3). ==:
x^3-6x^2+12x=14+6x-12. ==:
Ans. Is 2

15. May 31, 2016

### Chaos_Enlightened

Thank You All!!!!!!!

16. May 31, 2016

### Sahil Kukreja

Your question is wrong. it should be :- x= 2 + 2^(1/3) + 2^(2/3)

otherwise your calculations are wrong, you can just check.

17. May 31, 2016

### SammyS

Staff Emeritus
Hello Chaos_Enlightened. Welcome to PF !

Your original expression, has some ambiguity in it.

Using the standard Order of Operations, What you wrote, x=2+2^1/2+2^2/3 , literally means $\ x=2+2^1/2+2^2/3\ .$

Use parentheses around the fractional exponents as in x=2+2^(1/2)+2^(2/3), which gives $\ x=2+2^{1/2}+2^{2/3}\ .$

You may find the superscript feature (X2 button) helpful as well. Using this gives: x=2+21/2+22/3.

LaTeX was used for these.

18. May 31, 2016

### Chaos_Enlightened

Thank you Sammy's

19. May 31, 2016

### late347

I dont think that it factorizes any more than that at least in integers. I could be wrong though it would not be the first time...

I think it seems that the first common and therefore true factor is the lonely (x) (something...)

(x) (x2 -6x +6)

closest I managed to get was (x-3)(x-3) which yields out too much stuff. x^2 -6x + 9

(x-2) (x-3) yields too little stuff x^2 -5x + 6

x* x * (x-6 + 6/x) = (x^3) - (6x^2) +( 6x)

20. Jun 1, 2016

### late347

for x2 -6x +6 = 0

x is either (6+ √12) / 2
or
x is (6-√12) / 2

Its not very easy to calculate the original equation when the tip is used to switch 2 to the left side and cube both sides. Newton's binomial formula or pascals triangle can be used here. Maybe I am on the wrong track but looking at the calculation it will not be easy or fast to compute those roots into cubes. (X-2)^3 will be easier but also quite tedious.
x= 2 + 20.5 + 22/3
(x-2)3 = [(√2) + 3√4) ] 3

(x-2)^3 = [x+(-2)]3
calculate with binomial theorem or Newton's binomial formula which is found in some mathematics formula books ( It seems I used simplified version from wikipedia https://en.wikipedia.org/wiki/Binomial_theorem scroll down in the article until there is (x+y)3). At least the factors themselves can be seen from pascal's triangle easily. In this case the factors are 1,3,3, and 1.

x3-6x2 +12x -8

I'm sure that the right hand side yields something even uglier with pen-and--paper calculation...