Question

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Answer 1

Let $\text{x}-\dfrac{1}{\text{x}} = \text{q}$

Cubing on both sides of the equation,

$\bigg(\text{x} - \dfrac{1}{\text{x}}\bigg)^3 = \text{q}^3\\\\\text{x}^3 - \dfrac{1}{\text{x}^3} - 3\bigg(\text{x} - \dfrac{1}{\text{x}}\bigg) = \text{q}^3$

$(14) - 3(\text{q}) = \text{q}^3 \:\:\:\:\:\bigg[\text{Given that }\text{x}^3 - \dfrac{1}{\text{x}^3} = 14\bigg]$

$\text{q}^3 + 3\text{q} - 14 = 0$

$(\text{q} - 2)$ is a factor of $(\text{q}^3 + 3\text{q} - 14)$ as $\text{q}(2) = 0$.   [According to remainder theorem]

Therefore the other factor of $(\text{q}^3 + 3\text{q} - 14)$ on dividing by $(\text{q} - 2)$ is $(\text{q}^2 + 2\text{q} + 7)$.

$\implies (\text{q} - 2)(\text{q}^2 + 2\text{q} + 7) = 0\\\implies \text{q} = 2$

The other values of q are not real numbers as $(\text{q}^2 + 2\text{q} + 7)$ can not be factored.

Therefore, $\bold{x - \dfrac{1}{x} = 2}$.