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Dahasolnce [82]
3 years ago
11

Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Mathematics
1 answer:
Shkiper50 [21]3 years ago
4 0

Answer:

By closure property of multiplication and addition of integers,

If x + \dfrac{1}{x} is an integer

∴ \left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right ) is an integer

From which we have;

x^3 + \dfrac{1}{x^3} 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;

x + \dfrac{1}{x}

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}

By simplification of the cube of the given integer expressions, we have;

\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )

Therefore, we have;

\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}

By rearranging, we get;

x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )

Given that  x + \dfrac{1}{x} is an integer, from the closure property, the product of two integers is always an integer, we have;

\left ( x + \dfrac{1}{x} \right) ^3 is an integer and 3\cdot \left (x + \dfrac{1}{x} \right ) is also an integer

Similarly the sum of two integers is always an integer, we have;

\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right  ) is an integer

\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  ) is an integer

From which we have;

x^3 + \dfrac{1}{x^3} is an integer.

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