# Show, that (7+5√2)^(1/3)+(7−5√2)^(1/3) is an integer

• MHB
• lfdahl
In summary, to prove that (7+5√2)^(1/3)+(7−5√2)^(1/3) is an integer, we can use the rational root theorem, which states that the numerator must be a factor of the constant term and the denominator must be a factor of the leading coefficient. Other methods for proving this include using the binomial theorem and the concept of conjugates. It is important to prove this as it showcases the elegance of the expression and has practical applications in physics and engineering. This proof can also be extended to similar expressions with the form (a+b√c)^(1/n)+(a−b√c)^(1/n).
lfdahl
Gold Member
MHB
Let $a = \sqrt[3]{7+5\sqrt{2}} + \sqrt[3]{7-5\sqrt{2}}$. Show (without the use of a calculator), that $a$ is an integer.

lfdahl said:
Let $a = \sqrt[3]{7+5\sqrt{2}} + \sqrt[3]{7-5\sqrt{2}}$. Show (without the use of a calculator), that $a$ is an integer.

$a = \sqrt[3]{7+5\sqrt{2}} + \sqrt[3]{7-5\sqrt{2}}$.
we have a is real
cube both sides to get using $(x+y)^3= x^3+y^3 + 3xy(x+y)$
$a^3= 7+5\sqrt{2} + 7-5\sqrt{2}+ 3 (\sqrt[3]{7+5\sqrt{2}})(\sqrt[3]{7-5\sqrt{2}})a$
$= 14 + 3a$
hence $a^3+3a-14=0$
by rational root theorem 2 is a root
hence $a^3-4a + 7a-14 =0$
or $(a-2)(a^2 + 2a) +7(a- 2) = (a-2)(a^2+2a+7)=0$
so a = 2 or $a^2+2a+7=0$ which gives complex roots
so a = 2 as a is real
now I leave some one to prove that 2 is integer

lfdahl said:
Let $a = \sqrt[3]{7+5\sqrt{2}} + \sqrt[3]{7-5\sqrt{2}}$. Show (without the use of a calculator), that $a$ is an integer.
My solution:
[sp]Expand $(1 \pm \sqrt2)^3$ binomially to get $$(1 \pm \sqrt2)^3 = 1 \pm3\sqrt2 + 6 \pm2\sqrt2 = 7 \pm 5\sqrt2.$$ So $1 \pm \sqrt2 = \sqrt[3]{7\pm 5\sqrt2}$, and $\sqrt[3]{7 + 5\sqrt2} + \sqrt[3]{7 - 5\sqrt2} = (1 + \sqrt2) + (1 - \sqrt2) = 2.$

[/sp]

$a = \sqrt[3]{7+5\sqrt{2}} + \sqrt[3]{7-5\sqrt{2}}$.
we have a is real
cube both sides to get using $(x+y)^3= x^3+y^3 + 3xy(x+y)$
$a^3= 7+5\sqrt{2} + 7-5\sqrt{2}+ 3 (\sqrt[3]{7+5\sqrt{2}})(\sqrt[3]{7-5\sqrt{2}})a$
$= 14 + 3a$
hence $a^3+3a-14=0$
by rational root theorem 2 is a root
hence $a^3-4a + 7a-14 =0$
or $(a-2)(a^2 + 2a) +7(a- 2) = (a-2)(a^2+2a+7)=0$
so a = 2 or $a^2+2a+7=0$ which gives complex roots
so a = 2 as a is real
now I leave some one to prove that 2 is integer

Well done, kaliprasad! A clear and sharp solution indeed! Thankyou for your participation!(Yes)

- - - Updated - - -

Opalg said:
My solution:
[sp]Expand $(1 \pm \sqrt2)^3$ binomially to get $$(1 \pm \sqrt2)^3 = 1 \pm3\sqrt2 + 6 \pm2\sqrt2 = 7 \pm 5\sqrt2.$$ So $1 \pm \sqrt2 = \sqrt[3]{7\pm 5\sqrt2}$, and $\sqrt[3]{7 + 5\sqrt2} + \sqrt[3]{7 - 5\sqrt2} = (1 + \sqrt2) + (1 - \sqrt2) = 2.$

[/sp]
This is a suprisingly short and elegant way to "crack the nut". Thankyou, Opalg, for your participation!

## 1. How can you prove that (7+5√2)^(1/3)+(7−5√2)^(1/3) is an integer?

To prove that (7+5√2)^(1/3)+(7−5√2)^(1/3) is an integer, we can use the rational root theorem. This theorem states that if a polynomial with integer coefficients has a rational root, then the numerator must be a factor of the constant term and the denominator must be a factor of the leading coefficient. Applying this theorem to our expression, we can see that the numerator (7+5√2) and the denominator (7) are both factors of the constant term (49) and the leading coefficient (1), respectively. Therefore, we can conclude that the expression is an integer.

## 2. Can you explain the concept of a rational root theorem?

The rational root theorem is a mathematical theorem that helps us determine if a polynomial with integer coefficients has a rational root. It states that if a polynomial has a rational root, then the numerator must be a factor of the constant term and the denominator must be a factor of the leading coefficient. This theorem is useful in proving that certain expressions, such as (7+5√2)^(1/3)+(7−5√2)^(1/3), are integers.

## 3. Are there any other methods for proving that (7+5√2)^(1/3)+(7−5√2)^(1/3) is an integer?

Yes, there are other methods for proving that (7+5√2)^(1/3)+(7−5√2)^(1/3) is an integer. One method is to use the binomial theorem, which states that (a+b)^n = Σ(n, k=0)(n choose k)a^(n-k)b^k, where (n choose k) represents the binomial coefficient. By applying this theorem to our expression, we can see that all the terms will cancel out except for the first and last terms, which will result in an integer. Another method is to use the concept of conjugates, which states that if a+b√c is a root of a polynomial with rational coefficients, then a−b√c must also be a root. By using this concept, we can show that (7+5√2)^(1/3) and (7−5√2)^(1/3) are conjugates, and their sum will always be an integer.

## 4. Why is it important to prove that (7+5√2)^(1/3)+(7−5√2)^(1/3) is an integer?

Proving that (7+5√2)^(1/3)+(7−5√2)^(1/3) is an integer not only shows the mathematical elegance and beauty of the expression, but it also has practical applications. For example, in physics and engineering, integers often represent discrete quantities, such as the number of atoms or particles in a system. Therefore, being able to prove that a certain expression is an integer can help us better understand and model real-world phenomena.

## 5. Can this proof be extended to similar expressions?

Yes, this proof can be extended to similar expressions that have the form (a+b√c)^(1/n)+(a−b√c)^(1/n), where a, b, and c are integers and n is a positive integer. This is because the rational root theorem, binomial theorem, and concept of conjugates can be applied to any expression with this form. However, the values of a, b, and c may need to be adjusted accordingly to satisfy the conditions of each theorem or concept.

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