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

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The discussion focuses on finding positive values of \( a \) such that the expression \( \sqrt[3]{3+\sqrt{a}}+\sqrt[3]{3-\sqrt{a}} \) results in an integer. Various solutions and methods are proposed, including algebraic manipulations and alternative approaches to simplify the expression. Participants share their findings and verify each other's solutions, emphasizing the importance of integer results. The conversation highlights the mathematical principles involved in solving the problem. Ultimately, the goal is to identify all possible positive values of \( a \) that satisfy the condition.
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Determine the positive numbers, $a$, such that the sum:

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

is an integer.
 
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lfdahl said:
Determine the positive numbers, $a$, such that the sum:

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

is an integer.
my solution:
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
 
Last edited:
Albert said:
my solution:
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

Would you please show the other $a$ solution?
 
lfdahl said:
Would you please show the other $a$ solution?
another solution:
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 ($\sqrt[3]{3+\sqrt a}=x$)
($\sqrt[3]{3-\sqrt a}=y$)
 
Last edited:
Albert said:
another solution:
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:

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 $2$:$n = 1$: $x = \frac{368}{27} = 13,\overline{629}$.$n = 2$: $x = \frac{242}{27} = 8,\overline{962}$.
 
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