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

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The discussion centers on proving that the sum of the cube roots of two irrational numbers, specifically $\sqrt[3]{45+29\sqrt{2}} + \sqrt[3]{45-29\sqrt{2}}$, is rational. Participants express admiration for MarkFL's proof, indicating it is well-received and considered insightful. The conversation highlights the effectiveness of the approaches taken by MarkFL and mathbalarka, with some contributors feeling their own solutions are less impressive in comparison. Overall, the consensus is that the sum is indeed rational, supported by the presented proofs. The discussion showcases a collaborative effort to understand the properties of cube roots in relation to rationality.
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prove that $\sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}} $ is rational
 
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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

$$\sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}} = 6 \;\;\; \blacksquare$$
 
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kaliprasad said:
prove that $\sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}} $ is rational

$$\sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}}=\sqrt[3]{(3+\sqrt{2})^3}+\sqrt[3]{(3-\sqrt{2})^3}=6$$
 
MarkFL said:
$$\sqrt[3]{45+29\sqrt{2}}+ \sqrt[3]{45-29\sqrt{2}}=\sqrt[3]{(3+\sqrt{2})^3}+\sqrt[3]{(3-\sqrt{2})^3}=6$$

Aww...that's brilliant, MarkFL!(Yes):cool:
 
That deserves more thanks. My answer looks puny compared to that (Tmi)
 
anemone said:
Aww...that's brilliant, MarkFL!(Yes):cool:

I have a small confession to make:

I used a CAS to solve the system:

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

$$45-29\sqrt{2}=(a-b\sqrt{2})^3$$
 
hats of to markfl for a good ans.

my solution is almost same as mathbalarka
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
 
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