How to Solve ∛(7+5√2) - ∛(5√2-7) for Junior Olympiad?

[ranga519]

Hi,I'm new to this website.
This a question i came across in a junior Olympiad paper.Need help solving it.Thanks

∛(7+5√2) - ∛(5√2-7)

 

[Greg]

Hi ranga519 and welcome to MHB! :D

How can we check if $5\sqrt2+7$ is a cube? That is, if there is a number that, when cubed, is equal to $5\sqrt2+7$, how can we find it?

 

[MarkFL]

Hi,I'm new to this website.
This a question i came across in a junior Olympiad paper.Need help solving it.Thanks

∛(7+5√2) - ∛(5√2-7)

I would take Greg's advice and write:

\(\displaystyle 5\sqrt{2}\pm7=(b\sqrt{2}\pm a)^3=\pm a^3+3\sqrt{2}a^2b\pm6ab^2+2\sqrt{2}b^3=\left(3a^2b+2b^3\right)\sqrt{2}\pm\left(a^3+6ab^2\right)\)

We can see by inspection that one solution is:

\(\displaystyle (a,b)=(1,1)\)

Hence:

\(\displaystyle \sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{(\sqrt{2}+1)^3}-\sqrt[3]{(\sqrt{2}-1)^3}=(\sqrt{2}+1)-(\sqrt{2}-1)=2\)

 

[MountEvariste]

Let $u = \sqrt[3]{7+5\sqrt{2}}$ and $v= \sqrt[3]{7-5\sqrt{2}}$. We're interested in the sum $u+v$. First, observe that $u \cdot v = -1$. Also, $(u+v)^3 = u^3+3uv (u+v)+v^3$ and since $u \cdot v = -1$ and $u^3+v^3 = 14$ we have $(u+v)^3 = 14-3(u+v).$ Now let $x=u+v$. It now becomes solving the polynomial $x^3+3x-14 = 0$. By inspection, $x=2$ satisfies the equation, and thus we can write it as $(x-2)(x^2+2x+7)=0$, and we easily see that it has no other real roots. Hence $x=2$ is the only real root; hence $u+v=2$.

 

[ranga519]

Thank you very much for the help,
I find this website really helpful.