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## Homework Statement

If x=2+2^1/2+2^2/3. Then x^3-6x^2+6x=?

## Homework Equations

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

## The Attempt at a Solution

x^2= 6+3(2^2/3)+2^4/3+4(2^1/3)[/B]

- Thread starter Chaos_Enlightened
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- #1

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If x=2+2^1/2+2^2/3. Then x^3-6x^2+6x=?

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

x^2= 6+3(2^2/3)+2^4/3+4(2^1/3)[/B]

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Should I go on and find x^3

- #3

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

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Ok

- #5

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

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I think there is a calculation mistake here, sorry if i am wrong.Got x^3-6x^2+12x=14+6(2^1/3+2^2/3)

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- #7

Nidum

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

What do you do then ?

What do you do then ?

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- #8

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The question is solvable without calculating square roots and cube roots.Hint : work out the numerical value of x from the first equation .

What do you do then ?

- #9

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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....

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

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- #10

Nidum

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Think about what sort of answer the question really requires .

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That might just do it

- #12

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Which one should I factorise

- #13

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- #14

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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

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Thank You All!!!!!!!

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otherwise your calculations are wrong, you can just check.

- #17

SammyS

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Hello Chaos_Enlightened. Welcome to PF !## Homework Statement

If x=2+2^1/2+2^2/3. Then x^3-6x^2+6x=?

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 (X

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Thank you Sammy's

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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) (x

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

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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 + 2

(x-2)

(x-2)^3 = [x+(-2)]

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)

x

I'm sure that the right hand side yields

- #21

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*I recommend you start from the left side first (x-2)*^{3}

*I tend to agree with Sahil Kukreja's advice.*He or she provided a good tip. ( I think ) .- I think I got the basic idea, in order to get the "rough answer" in unfinished form. It should still be mathematically correct but your teacher will likely have the finished form answer in the answerbook
- try to get the "rough form answer" first, then worry later about expanding the (squareroot + cubic root)
^{3}

- Mathematically it's not very significant -
*it's more of a style issue, I suppose.*

(x-2)

. In the end phase we must give the answer for the question "What does this clause yield: clause is; x

The answer must be another clause, which equals that quadratic clause. The full clause in the ending (both sides of equation) will be an equation, in that sense. So, therefore the ending clause should be properly balanced like all equations, if that makes it clear. This is my interpretation at least.

But the question asks simply (what does the left side clause give you, on the right side clause?)

- We know for a fact that this equation holds true

- Therefore

- Therefore

to-do-list

1. first equation was x= 2 + 2

2. in the first equation, minus 2 from both sides.

3. in the first equation, then cube both sides (third power both sides)

4. calculate the (x-2)

5. according to my limited knowledge, you can substitute the negative sign, simply by using brackets and minus inside

(x-2)^3 = [x+(-2)] ^3

When you calculate

a+(-b)

it is the same as

a-b

At least I hope so because otherwise I screwed up my own solution hehe...

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Then I plugged my paper results into the calculator and verified the equation holds true on both sides.

I think that it may be unnecessary to expand the (2

But one method is to use the binomial formula for (x+y)

x

Calculator can be used at the end to verify that both sides of equation hold true. That's what I would do, but I could be wrong though.

- #23

SammyS

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Where did you get ##\ 6\left(2^{1/3}+2^{2/3}\right) \ ?##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

... or should the original problem state that ##\ x=2+2^{1/3}+2^{2/3}\ ## rather than ##\ x=2+2^{1/2}+2^{2/3}\ ? ##

- #24

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Yes, thats what i was saying in post #16.Where did you get ##\ 6\left(2^{1/3}+2^{2/3}\right) \ ?##

... or should the original problem state that ##\ x=2+2^{1/3}+2^{2/3}\ ## rather than ##\ x=2+2^{1/2}+2^{2/3}\ ? ##

- #25

SammyS

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Post #16:Yes, that's what i was saying in post #16.

I see that I missed that post of yours.

otherwise your calculations are wrong, you can just check.

Then it's questionable as to whether OP will return to tell us. He's had over a day to respond to your post.

Certainly, ##\ x=2+2^{1/3}+2^{2/3}\ ## is much easier to work with. However, I was surprised that the problem as posted (likely with a typo) gave a fairly simple result, having only two terms.

##\ 6\cdot 2^{5/6}-4\cdot2^{1/2}=5.033\dots\ ##

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