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$$\sqrt[3]{1+\sqrt{x}}=\sqrt{1+\sqrt[3]{x}}$$

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Jomo said: I did not want to bother to raise each side to the 6th power and work through this mess. However I do understand your point about deriving the solutions. Click to expand... You don't have to do the heavy lifting on your own, you can use readily available software to do the fiddly algebra. ... or of you are lucky enough to have them, task one of your graduate students to do it.
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CaptainBlack said: However the point of this being a puzzle is to derive the solutions, that is if these are the only solutions to prove it. Click to expand... I did not want to bother to raise each side to the 6th power and work through this mess. However I do understand your point about deriving the solutions.
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Spoiler I raised both sides to the power of 6. This gave a squared expression equal to a cubed expression. Expand it out and things cancel. Assuming x is not 0, this leads to a quadratic with no real solutions. But x=0 is an obvious solution, so is the only real solution. (I'm being a bit lazy not LaTexing it all out! )
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Jomo said: Actually, I bet x=0 is the only real zero. Click to expand... There is another "obvious" solution in the extended reals at $+\infty$ (which might be why I asked about the domain?) However the point of this being a puzzle is to derive the solutions, that is if these are the only solutions to prove it.
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Actually, I bet x=0 is the only real zero.
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x=0 is one obvious solution.
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CaptainBlack said: What are we expected to do with this? Solve it? In what domain? Click to expand... This problem and two methods for its solutions came off of the SyberMath site from one day ago.
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CaptainBlack said: What are we expected to do with this? Solve it? In what domain? Click to expand... Sorry, the direction is: Find all values of $x$ that satisfy the equation over the reals, but you're welcome to go for complex as well.
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SupremeCookie said: $$\sqrt[3]{1+\sqrt{x}}=\sqrt{1+\sqrt[3]{x}}$$ Click to expand... What are we expected to do with this? Solve it? In what domain?
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