Is the sum of two cube roots of irrational numbers rational?

[kaliprasad]

prove that $\sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}} $ is rational

 

[mathbalarka]

Spoiler

Let $x = \sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}}$. Then by application of binomial theorem for $n = 3$,

$\boxed{x^3} = \left ( \sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}} \right )^3$

$=\left(\sqrt[3]{45+29\sqrt{2}}\right)^3+ \left(\sqrt[3]{45-29\sqrt{2}}\right)^3$ $ + 3 \left (\sqrt[3]{45+29\sqrt{2}} \right) \left(\sqrt[3]{45-29\sqrt{2}}\right) \underbrace{\left (\sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}} \right )}_{=x}$
$= 45 + 29\sqrt{2} + 45 - 29\sqrt{2} + 3 \cdot \underbrace{\sqrt[3]{45^2 - 2 \cdot 29^2}}_{=7} \cdot x = \boxed{90 + 21x}$

Thus, the given expression is a root of $x^3 - 21x - 90 = 0$. By the rational root theorem, if this has a rational root, it must also be an integer. Thus looking for integer factors we find that $x^3 - 21x - 90 = (x - 6)(x^2 + 6x + 15)$. The quadratic factor has discriminant $\Delta = 6^2 - 4 \cdot 15 = -24$ and hence all of the roots complex. Thus, $6$ is the only rational (well, integer) root of the cubic, forcing

$$\sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}} = 6 \;\;\; \blacksquare$$

 

[MarkFL]

prove that $\sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}} $ is rational

Spoiler

\(\displaystyle \sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}}=\sqrt[3]{(3+\sqrt{2})^3}+\sqrt[3]{(3-\sqrt{2})^3}=6\)

 

[anemone]

Spoiler

\(\displaystyle \sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}}=\sqrt[3]{(3+\sqrt{2})^3}+\sqrt[3]{(3-\sqrt{2})^3}=6\)

Aww...that's brilliant, MarkFL!(Yes)

 

[mathbalarka]

That deserves more thanks. My answer looks puny compared to that (Tmi)

 

[MarkFL]

Aww...that's brilliant, MarkFL!(Yes)

I have a small confession to make:

Spoiler

I used a CAS to solve the system:

\(\displaystyle 45+29\sqrt{2}=(a+b\sqrt{2})^3\)

\(\displaystyle 45-29\sqrt{2}=(a-b\sqrt{2})^3\)

 

[kaliprasad]

hats of to markfl for a good ans.

my solution is almost same as mathbalarka

Spoiler

let
$x=\sqrt[3]{45 + 29\sqrt2} + \sqrt[3]{45 - 29\sqrt2}$
or
$x-\sqrt[3]{45 + 29\sqrt2} - \sqrt[3]{45 - 29\sqrt2}= 0$
using $a+b+c = 0 => a^3+b^3+ c^3 = 3abc$
we get
$x^3-(45 + 29\sqrt2) - (45 - 29\sqrt2)= 3 x\sqrt[3]{(45 + 29\sqrt2)(45 - 29\sqrt2)}$
or
$x^3-90= 3x\sqrt[3]{45^2 - 2 * 29^2})$
or $x^3-90 = 21x$
or $x^3 - 21x - 90 = 0$
or $(x-6)(x^2+6x+ 15) = 0$
has one real root = 6 and 2 complex roots
hence given expression = 6