Simplifying Radical Expression

Question

I came across this question in a text book while preparing for return to university. It has been a few years since I flexed the old grey matter and I have not been able to arrive at the solution. Any help would be much appreciated. The question is:

Express the following in the form $ x + y \sqrt{2}$ with x and y rational numbers: $\sqrt[3]{(7+5\sqrt{2})}$

B.O.B gives the answer as:

$1+\sqrt{2}$

When this expression is cubed it does equal the original expression but I am unsure how to arrive at the solution.

Answer 1

Hint: Write $\sqrt[3]{(7+5\sqrt{2})}=x + y \sqrt{2}$, then raise both sides to the third power and solve for $x$ and $y$. You will get:

$x(x^2+6y^2)=7$

$y(3x^2+2y^2)=5$

Then eliminate one of the variables (say $y$) and you will get a polynomial $P(x)=0$. Then you have to look for the rational roots of the polynomial (if such roots exist) with Rational Root Theorem

Of course in this particular case, you might just observe that $x=y=1$ is solution of the system (which is strongly suggested by the fact that both $5$ and $7$ are prime and both $x^2+6y^2$ and $3x^2+2y^2$ are positive).

Answer 2

The problem is that not every expression of the form $\sqrt[3]{a+b\sqrt2}\quad$can be expressed in the form $c+d\sqrt2.\quad\{a,b,c,d\} \in \text{integers}$

I have a feeling that whoever posed the question to you already knew that a solution existed.