(3+√a)^(1/3)+(3-√a)^(1/3) is an integer, find a

[lfdahl]

Determine the positive numbers, $a$, such that the sum:

$$\sqrt[3]{3+\sqrt{a}}+\sqrt[3]{3-\sqrt{a}}$$

is an integer.

 

[Albert1]

Determine the positive numbers, $a$, such that the sum:

$$\sqrt[3]{3+\sqrt{a}}+\sqrt[3]{3-\sqrt{a}}$$

is an integer.

my solution:

Spoiler

let $\sqrt[3]{3+\sqrt{a}}=x$
$\sqrt[3]{3-\sqrt{a}}=y$
consider $x>0,y>0$
$x^3+y^3=(x+y)(x^2-xy+y^2)=6=2\times 3=3\times 2=6\times 1=1\times 6$
now $x+y=2---(1)$
and $x^2-xy+y^2=3$ $\rightarrow xy=\dfrac {1}{3}---(2)$
satisfy the condition $a>0$
$x,y$ are roots of $3t^2-6t+1=0$
$x=\dfrac{3+\sqrt 6}{3}$
$y=\dfrac{3-\sqrt 6}{3}$
and we get :$a\approx 8.963$
if $x+y<0$
$x^3+y^3=(x+y)(x^2-xy+y^2)=6=-2\times -3=-3\times -2=-6\times -1=-1\times -6$
with the same method we may get another "a" or no solution

 

[lfdahl]

my solution:

Spoiler

let $\sqrt[3]{3+\sqrt{a}}=x$
$\sqrt[3]{3-\sqrt{a}}=y$
consider $x>0,y>0$
$x^3+y^3=(x+y)(x^2-xy+y^2)=6=2\times 3=3\times 2=6\times 1=1\times 6$
only $x+y=2---(1)$
and $x^2-xy+y^2=3$ $\rightarrow xy=\dfrac {1}{3}---(2)$
satisfy the condition $a>0$
$x,y$ are roots of $3t^2-6t+1=0$
$x=\dfrac{3+\sqrt 6}{3}$
$y=\dfrac{3-\sqrt 6}{3}$
and we get :$a\approx 8.963$
if $x+y<0$
$x^3+y^3=(x+y)(x^2-xy+y^2)=6=-2\times -3=-3\times -2=-6\times -1=-1\times -6$
with the same method we may get another "a" or no solution

Spoiler

Would you please show the other $a$ solution?

 

[Albert1]

Spoiler

Would you please show the other $a$ solution?

another solution:

Spoiler

sorry another solution will happen at :
$x+y=1$ and
$x^2-xy+y^2=6,or \,\, 3xy=-5$
which gives $x=1.8844,or\, -0.8844$
both yield $a=13.63$
here ($\sqrt[3]{3+\sqrt a}=x$)
($\sqrt[3]{3-\sqrt a}=y$)

 

[lfdahl]

another solution:

Spoiler

sorry another solution will happen at :
$x+y=1$ and
$x^2-xy+y^2=6,or \,\, 3xy=-5$
which gives $x=1.8844,or\, -0.8844$
both yield $a=13.63$
here ($\sqrt[3]{3+\sqrt a}=x$)
($\sqrt[3]{3-\sqrt a}=y$)

Well done, thankyou Albert!An alternative approach can be found here:

Spoiler

Let \[n = \sqrt[3]{3+\sqrt{x}}+\sqrt[3]{3-\sqrt{x}}, \: \: x >0.\]

Then

\[n^3 = 6+3n\left [ (3+\sqrt{x})(3-\sqrt{x})\right ]^{1/3}\]

Hence, \[\left ( \frac{n^3-6}{3n} \right )^3 = 9-x \Rightarrow x = 9 - \left ( \frac{n^3-6}{3n} \right )^3 > 0.\]

The term $\left ( \frac{n^3-6}{3n} \right )^3$ is monotone increasing and larger than $9$ for $n \ge 3$, so

it suffices to let $n$ be $1$ or $2$:$n = 1$: $x = \frac{368}{27} = 13,\overline{629}$.$n = 2$: $x = \frac{242}{27} = 8,\overline{962}$.