Respuesta :
Answer:
By closure property of multiplication and addition of integers,
If [tex]x + \dfrac{1}{x}[/tex] is an integer
∴ [tex]\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )[/tex] is an integer
From which we have;
[tex]x^3 + \dfrac{1}{x^3}[/tex] is an integer
Step-by-step explanation:
The given expression for the positive integer is x + x⁻¹
The given expression can be written as follows;
[tex]x + \dfrac{1}{x}[/tex]
By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;
[tex]\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}[/tex]
By simplification of the cube of the given integer expressions, we have;
[tex]\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )[/tex]
Therefore, we have;
[tex]\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}[/tex]
By rearranging, we get;
[tex]x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )[/tex]
Given that [tex]x + \dfrac{1}{x}[/tex] is an integer, from the closure property, the product of two integers is always an integer, we have;
[tex]\left ( x + \dfrac{1}{x} \right) ^3[/tex] is an integer and [tex]3\cdot \left (x + \dfrac{1}{x} \right )[/tex] is also an integer
Similarly the sum of two integers is always an integer, we have;
[tex]\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )[/tex] is an integer
[tex]\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )[/tex] is an integer
From which we have;
[tex]x^3 + \dfrac{1}{x^3}[/tex] is an integer.
