Is the given quantity an integer?

[anemone]

Your calculator tells you that the quantity $y=(2\cdot\sqrt[3]{2}+1-\sqrt{12\cdot\sqrt[3]{2}-15})^3$ is approximately an integer. Is $y$ exactly an integer? Justify your answer.

 

[kaliprasad]

Your calculator tells you that the quantity $y=(2\cdot\sqrt[3]{2}+1-\sqrt{12\cdot\sqrt[3]{2}-15})^3$ is approximately an integer. Is $y$ exactly an integer? Justify your answer.

Spoiler

yes it is 32

as

Spoiler

Because under the square root we have $\sqrt[3]{2}$
It suggests that the square root is
a + b $\sqrt[3]{2}$ + c $\sqrt[3]{4}$
And taking square we get
$a^2+ b^2\sqrt[3]{4} +2c^2\sqrt[3]{4} +2ab\sqrt[3]{2} + 4bc + 2ac\sqrt[3]{4} = 12\sqrt[3]{2}−15$
Equating the parts that is rational , $\sqrt[3]{2}$ and $\sqrt[3]{4}$
We get
$a^2 + 4bc = - 15$
$b^2 + 2ac = 0$
$c^2 + ab = 6$
Solving these ( I do not know how to solve but guess work a = 1, b= 2, c =- 2
so square root = 1 + 2 $\sqrt[3]{2}$ - 2 $\sqrt[3]{4}$

so given expression is 2 $\sqrt[3]{4}$ which is cubed to give 32

 

[Albert1]

Your calculator tells you that the quantity $y=(2\cdot\sqrt[3]{2}+1-\sqrt{12\cdot\sqrt[3]{2}-15})^3$ is approximately an integer. Is $y$ exactly an integer? Justify your answer.

Spoiler

let $\sqrt[3]{2}=x$
we have :$x^3=2---(1)$
$y= (2x+1-\sqrt {12x-15})^3$
if y is an integer then we may suggest :
$\sqrt {12x-5}=2x+1-ax^2---(2)$
(here a is an integer)
square both sides of (2) and use of (1) we get a=2
$\therefore y=(2x^2)^3=32$

 

[anemone]

Thank you all for participating and sorry for the late reply.

Suggested solution by other:

Spoiler

According to the calculator, $y$ is approximately equal to 32, and so we would like to decide whether or not the quantity $x=2\cdot\sqrt[3]{2}+1-\sqrt{12\cdot\sqrt[3]{2}-15}$ is exactly equal to $\sqrt[3]{32}$.

For notational simplicity, let $a=\sqrt[3]{2}$ so that $x=2a+1-\sqrt{12a-15}$.

Also, since $32=2^5$, we see that $\sqrt[3]{32}=2a^2$ and hence we need to determine whether or not the equation $\sqrt{12a-15}=2a+1-2a^2$ is valid.

Using calculator, we see that the right side of this equation exceeds 0.3, and so it is definitely positive. We can thus check the equation by showing that the squares of both sides are equal. If we square the right side and combine like terms, we get $4a^4-8a^3+4a+1$. But since $a^3=2$, we see that $a^4=2a$ and thus $4a^4-8a^3+4a+1=12a-15$. Consequently, our equation is true, and $y$ is indeed exactly equal to 32.

 

[CellCoree]

To determine if the given quantity $y$ is exactly an integer, we must first understand what an integer is. An integer is a whole number, meaning it does not have any fractional or decimal parts. It can be positive, negative, or zero.

Looking at the expression for $y$, we can see that it involves a combination of square roots and cube roots, which are irrational numbers. This means that the exact value of $y$ cannot be expressed as a simple fraction or decimal. However, our calculator is able to approximate the value of $y$ as an integer.

Therefore, while $y$ may appear to be an integer on our calculator, it is not exactly an integer. It is only an approximation of an integer. This is because the calculator has a limited number of digits and must round off the value of $y$ to display it, leading to a small margin of error.

In conclusion, $y$ is not exactly an integer, but it is very close to being one. It is important to note that this is not a flaw or error in the calculator, but rather a limitation of representing irrational numbers in a finite number of digits.