<![CDATA[Math Is Fun Forum / Express ∛(7 + 5√2) in the form x + y√2]]> 2019-01-11T10:08:33Z FluxBB http://www.mathisfunforum.com/viewtopic.php?id=24737 <![CDATA[Re: Express ∛(7 + 5√2) in the form x + y√2]]> Hi zetafunc,

I see where you're going with this, expanding (x + y√2)^3 gives x^3 + 3√2x^2y + 6xy^2 + 2√2y^3 = 7 + 5√2

Now if I substitute x = 1 and y = 1 on the LHS I do get the RHS, and this gives 1 + √2 as required,  but is there a more systematic way to find the values of x and y?
It reminds me of the technique of using undetermined coefficients in partial fractions, but not sure how to do it in this case...
Thanks again.

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http://www.mathisfunforum.com/profile.php?id=219882 2019-01-11T10:08:33Z http://www.mathisfunforum.com/viewtopic.php?pid=406705#p406705
<![CDATA[Re: Express ∛(7 + 5√2) in the form x + y√2]]> Hi segfault,

Welcome to the forum!

Suppose that there are some values x and y for which ∛(7 + 5√2) = x + y√2. What happens if you cube both sides of that equation?

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http://www.mathisfunforum.com/profile.php?id=206089 2019-01-10T19:39:45Z http://www.mathisfunforum.com/viewtopic.php?pid=406699#p406699
<![CDATA[Express ∛(7 + 5√2) in the form x + y√2]]> The expression given is ∛(7 + 5√2), which is to be expressed in the form x + y√2. The answer given in the back of the book is 1 + √2, which is indeed numerically the same as ∛(7 + 5√2), but I'm damned if I can see how you get from one to the other. Any suggestions? Thanks in advance!

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http://www.mathisfunforum.com/profile.php?id=219882 2019-01-10T17:36:17Z http://www.mathisfunforum.com/viewtopic.php?pid=406698#p406698