If $x^4 + \dfrac{1}{x^4} = 119$, x > 1, then find the value of $x^3 - \dfrac{1}{x^3}$

Given,

$x^4 + \dfrac{1}{x^4} = 119$, and x > 1.

$x^2 + \dfrac{1}{x^2}$

Using identity

$\Big(x^2 + \dfrac{1}{x^2}\Big)^2 = x^4 + \dfrac{1}{x^4} + 2$

Substitute the given value:

$\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = 119 + 2 \\[1em] \Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 = 121 \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \sqrt{121} \\[1em] \Rightarrow x^2 + \dfrac{1}{x^2} = \pm 11$

Since $x > 1$, $x^2 + \dfrac{1}{x^2}$ must be positive.

So, $x^2 + \dfrac{1}{x^2} = 11$

$x - \dfrac{1}{x}$

Using identity

$\Big(x - \dfrac{1}{x}\Big)^2 = x^2 + \dfrac{1}{x^2} - 2$

Substitute the value:

$\Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 11 - 2 \\[1em] \Rightarrow \Big(x - \dfrac{1}{x}\Big)^2 = 9 \\[1em] \Rightarrow x - \dfrac{1}{x} = \pm\sqrt{9} \\[1em] \Rightarrow x - \dfrac{1}{x} = \pm 3$

Since $x > 1$, $x$ is greater than $\dfrac{1}{x}$, so $x - \dfrac{1}{x}$ must be positive.

So, $x - \dfrac{1}{x} = 3$

$x^3 - \dfrac{1}{x^3}$

By using the identity:

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Let $a = x$ and $b = \dfrac{1}{x}$.

$\Rightarrow x^3 - \dfrac{1}{x^3} = \Big(x - \dfrac{1}{x}\Big)\Big(x^2 + x \cdot \dfrac{1}{x} + \dfrac{1}{x^2}\Big) \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = \Big(x - \dfrac{1}{x}\Big)\Big(x^2 + \dfrac{1}{x^2} + 1\Big)\\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = (3)(11 + 1) \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = (3)(12) \\[1em] \Rightarrow x^3 - \dfrac{1}{x^3} = 36$

Hence, $x^3 - \dfrac{1}{x^3} = 36$.