Solve Radical Equation: -4/9(x^9)[(3/x^2)-(1/(2^1/3))] when x = 2^1/6

[DecayProduct]

Homework Statement

Sorry, I don't know latex yet! The problem: -4/9(x^9)[(3/x^2)-(1/(2^1/3))] when x = 2^1/6. I'll put it in words too, cause that looks confusing. Negative four-ninths x to the ninth, times three over x squared minus one over the cube root of two, when x equals the sixth root of two.

Homework Equations

x^9 = (2^1/6)^9 = 2^9/6 = 2^3/2
x^2 = (2^1/6)^2 = 2^2/6 = 2^1/3
Rationalizing the denominators for the second part of the equation, I get 3/x^2 = 3/2^1/3 = 3(2^2/3)/2 and 1/(2^1/3) = (2^2/3)/2

The Attempt at a Solution

Now, when I put them all together I get -4/9(2^3/2)(2^2/3) = -4/9(2)(2^1/6) = -8/9(2^1/6). Now here's my problem, my book answers the problem as -16/9(2^1/6). I've done it over and over again, and still I get the -8/9. I have also had more than one wrong answer from the book, but this one I'm just not sure. All I want is someone else to work the problem and tell me whether I'm right or where I went wrong.

 

[berkeman]

Homework Statement

Sorry, I don't know latex yet! The problem: -4/9(x^9)[(3/x^2)-(1/(2^1/3))] when x = 2^1/6. I'll put it in words too, cause that looks confusing. Negative four-ninths x to the ninth, times three over x squared minus one over the cube root of two, when x equals the sixth root of two.

Homework Equations

x^9 = (2^1/6)^9 = 2^9/6 = 2^3/2
x^2 = (2^1/6)^2 = 2^2/6 = 2^1/3
Rationalizing the denominators for the second part of the equation, I get 3/x^2 = 3/2^1/3 = 3(2^2/3)/2 and 1/(2^1/3) = (2^2/3)/2

The Attempt at a Solution

Now, when I put them all together I get -4/9(2^3/2)(2^2/3) = -4/9(2)(2^1/6) = -8/9(2^1/6). Now here's my problem, my book answers the problem as -16/9(2^1/6). I've done it over and over again, and still I get the -8/9. I have also had more than one wrong answer from the book, but this one I'm just not sure. All I want is someone else to work the problem and tell me whether I'm right or where I went wrong.

There's a tutorial for LaTex in the Learning Materials forum, and the first post in it has some good crib sheets to get you started:

https://www.physicsforums.com/showthread.php?t=8997

Also, when you start a new thread or use Advanced Reply, there is a little \Sigma button in the upper right of the window. Clicking on that brings up a LaTex selection box over on the right, which you can use to post your LaTex equations.

On your question, first of all you can check your answer and the book's answer by just using a calculator. Use the calculator to find what the 1st formula is equal to, given that x = 2^\frac{1}{6}. Do you get your answer or the book's answer?

Also, I've bolded the part of your work that doesn't look right to me. Look again at the equation -- do you really need to do much work to get the two parts over a common denominator...?

 

[DecayProduct]

Thanks for the reply. I found where I was making a mistake. At the very end, when I ended up with -4/9(2^13/6), I kept reducing it to -4/9(2)(2^1/6). When in fact, it should be -4/9(4)(2^1/6) DUH! I must have done it 100 times and made the same stupid mistake every time!

And you are right about the common denominators, but I have been taught to rationalize every denominator, even if they are the same. And I just didn't realize it. Thanks!