Proof that $N = {(45 + 29 \sqrt{2})}^{\frac{1}{3}} + {(45 - 29 \sqrt{2})}^{\frac{1}{3}}$ $N = (45+29 sqrt(2))^(1/3)+(45-29 sqrt(2))^(1/3)$ is a integer ?

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Proof that $N = {(45 + 29 \sqrt{2})}^{\frac{1}{3}} + {(45 - 29 \sqrt{2})}^{\frac{1}{3}}$ $N = (45+29 sqrt(2))^(1/3)+(45-29 sqrt(2))^(1/3)$ is a integer ?

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Answer 1 · 2016-09-13T22:14:36

Consider $t^{3} - 21 t - 90 = 0$ $t^3-21t-90 = 0$

This has one Real root which is $6$ $6$ a.k.a. ${(45 + 29 \sqrt{2})}^{\frac{1}{3}} + {(45 - 29 \sqrt{2})}^{\frac{1}{3}}$ $(45+29sqrt(2))^(1/3)+(45-29sqrt(2))^(1/3)$

Explanation:

Consider the equation:

$t^{3} - 21 t - 90 = 0$ $t^3-21t-90 = 0$

Using Cardano's method to solve it, let $t = u + v$ $t = u+v$

Then:

$u^{3} + v^{3} + 3 (u v - 7) (u + v) - 90 = 0$ $u^3+v^3+3(uv-7)(u+v)-90 = 0$

To eliminate the term in $(u + v)$ $(u+v)$ , add the constraint $u v = 7$ $uv=7$

Then:

$u^{3} + \frac{7^{3}}{u^{3}} - 90 = 0$ $u^3+7^3/u^3-90 = 0$

Multiply through by $u^{3}$ $u^3$ and rearrange to get the quadratic in $u^{3}$ $u^3$ :

${(u^{3})}^{2} - 90 (u^{3}) + 343 = 0$ $(u^3)^2-90(u^3)+343 = 0$

by the quadratic formula, this has roots:

$u^{3} = \frac{90 \pm \sqrt{90^{2} - (4 \cdot 343)}}{2}$ $u^3 = (90+-sqrt(90^2-(4*343)))/2$

$u^{3} = 45 \pm \frac{1}{2} \sqrt{8100 - 1372}$ $color(white)(u^3) = 45 +- 1/2sqrt(8100-1372)$

$u^{3} = 45 \pm \frac{1}{2} \sqrt{6728}$ $color(white)(u^3) = 45 +- 1/2sqrt(6728)$

$u^{3} = 45 \pm 29 \sqrt{2}$ $color(white)(u^3) = 45 +- 29sqrt(2)$

Since this is Real and the derivation was symmetric in $u$ $u$ and $v$ $v$ , we can use one of these roots for $u^{3}$ $u^3$ and the other for $v^{3}$ $v^3$ to deduce that the Real zero of $t^{3} - 21 t - 90$ $t^3-21t-90$ is:

$t_{1} = \sqrt[3]{45 + 29 \sqrt{2}} + \sqrt[3]{45 - 29 \sqrt{2}}$ $t_1 = root(3)(45+29sqrt(2))+root(3)(45-29sqrt(2))$

but we find:

${(6)}^{3} - 21 (6) - 90 = 216 - 126 - 90 = 0$ $(6)^3-21(6)-90 = 216 - 126 - 90 = 0$

So the Real zero of $t^{3} - 21 t - 90$ $t^3-21t-90$ is $6$ $6$

So $6 = \sqrt[3]{45 + 29 \sqrt{2}} + \sqrt[3]{45 - 29 \sqrt{2}}$ $6 = root(3)(45+29sqrt(2))+root(3)(45-29sqrt(2))$

$color(white)()$
Footnote

To find the cubic equation, I used Cardano's method backwards.

Answer 2 · 2016-09-13T23:07:29

$N = 6$ $N = 6$

Explanation:

Making $x = 45 + 29 \sqrt{2}$ $x = 45+29 sqrt(2)$ and $y = 45 - 29 \sqrt{2}$ $y = 45-29 sqrt(2)$ then

${(x^{\frac{1}{3}} + y^{\frac{1}{3}})}^{3} = x + 3 {(x y)}^{\frac{1}{3}} x^{\frac{1}{3}} + 3 {(x y)}^{\frac{1}{3}} y^{\frac{1}{3}} + y$ $(x^(1/3)+y^(1/3))^3=x + 3 (x y)^(1/3)x^(1/3)+3(x y)^(1/3) y^(1/3)+y$

${(x y)}^{\frac{1}{3}} = {(7^{3})}^{\frac{1}{3}} = 7$ $(x y)^(1/3) = (7^3)^(1/3) = 7$
$x + y = 2 \times 45$ $x+y =2 xx 45$

so

${(x^{\frac{1}{3}} + y^{\frac{1}{3}})}^{3} = 90 + 21 (x^{\frac{1}{3}} + y^{\frac{1}{3}})$ $(x^(1/3)+y^(1/3))^3 = 90 + 21(x^(1/3)+y^(1/3))$

or calling $z = x^{\frac{1}{3}} + y^{\frac{1}{3}}$ $z = x^(1/3)+y^(1/3)$ we have

$z^{3} - 21 z - 90 = 0$ $z^3-21 z-90 = 0$

with $90 = 2 \times 3^{2} \times 5$ $90 = 2 xx 3^2 xx 5$ and $z = 6$ $z = 6$ is a root so

$x^{\frac{1}{3}} + y^{\frac{1}{3}} = 6$ $x^(1/3)+y^(1/3) = 6$

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Proof that $N = {(45 + 29 \sqrt{2})}^{\frac{1}{3}} + {(45 - 29 \sqrt{2})}^{\frac{1}{3}}$ $N = (45+29 sqrt(2))^(1/3)+(45-29 sqrt(2))^(1/3)$ is a integer ?

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Proof that $N = {(45 + 29 \sqrt{2})}^{\frac{1}{3}} + {(45 - 29 \sqrt{2})}^{\frac{1}{3}}$ $N = (45+29 sqrt(2))^(1/3)+(45-29 sqrt(2))^(1/3)$ is a integer ?