How do I fix this surd inside a surd?

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To solve the surd inside a surd, start by defining x as 3√(2 - √3) + 3/√(2 - √3) and aim to prove that x² equals 54. Initial attempts using "=54" are incorrect since that is what needs to be proven. Simplification steps involve using algebraic identities to manipulate x², leading to cancellations and combining like terms. The solution was clarified when it was recognized that 54 simplifies to 3√6. Understanding these simplifications is crucial for resolving similar surd problems.
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Homework Statement
The question says to use the answer to a and b
Relevant Equations
I know I can rationalize. And a surd multiplied by itself removes the root
It’s the last one that I couldn’t do. I have tried the other two. And that’s ok.
You can see the attemp in the picture. I didn’t write it here.
Please help
 

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In your working you start with the "=54" there.
That's never going to work because that's essentially what you have to prove (since 54 is the square of ##3\sqrt 6##), so you can't use it.

Instead, define ##x = 3\sqrt{2 - \sqrt 3}+ \frac 3{ \sqrt{2 - \sqrt 3}}## and then try to prove that ##x^2=54##.
In the first step you can use (a) to simplify ##x^2##.
In the second step you can use (b) to simplify further.
You will have some cancellation and collecting of like terms to do.
You should be able to end up with 54.
 
andrewkirk said:
In your working you start with the "=54" there.
That's never going to work because that's essentially what you have to prove (since 54 is the square of ##3\sqrt 6##), so you can't use it.

Instead, define ##x = 3\sqrt{2 - \sqrt 3}+ \frac 3{ \sqrt{2 - \sqrt 3}}## and then try to prove that ##x^2=54##.
In the first step you can use (a) to simplify ##x^2##.
In the second step you can use (b) to simplify further.
You will have some cancellation and collecting of like terms to do.
You should be able to end up with 54.
Just figured it out. Forgot that 54 becomes 3root6
 
Thank you all
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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