If $$x^{4}+\frac{1}{x^{4}}=119,$$ then the value of $$x^{3}-$$ $$\frac{1}{x^{3}}$$ is

$$(x^{2}+\frac{1}{x}^{2})^2$$ = $$x^{4} + \frac{1}{x}^{4} + 2*x^{2}*\frac{1}{x}^{2}$$
= 119 +2
= 121
$$x^{2}+\frac{1}{x}^{2} = 11$$
$$(x - \frac{1}{x})^{2} = x^{2} + \frac{1}{x}^{2} - 2*x*\frac{1}{x}$$
= 11 -2
= 9
$$x - \frac{1}{x} = ±3$$
now $$(x-\frac{1}{x})^{3} = x^{3} - \frac{1}{x}^{3} -3*x*\frac{1}{x}(x -\frac{1}{x})$$

if x = 3 then $$x^{3} - \frac{1}{x}^{3} = 3^{3} + 3*3$$
= 27 +9
= 36

Similarly if x = -3 then $$x^{3} - \frac{1}{x}^{3} = (-3)^{3} + (-3)*3 = -27 - 9 = - 36$$