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
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.