If $$x^2$$ + 3x + 1 = 0, then the value of$$ x^{3}+\frac{1}{x^{3}}$$

$$x^2+3x+1=0$$

$$=$$> $$x^2=-3x-1$$

$$x^3\ +\ \ \frac{\ 1}{x^3}=\left(x+\ \frac{\ 1}{x}\right)\left(x^2+\ \frac{\ 1}{x^2}-1\right)$$

$$\ =\frac{\ x^2+1}{x}\left(x^2+\ \frac{\ 1}{x^2}-1\right)$$

$$\ =\frac{\ -3x-1+1}{x}\left(-3x-1+\ \frac{\ 1}{-3x-1}-1\right)$$

$$\ =\frac{\ -3x}{x}\left(-3x-2+\ \frac{\ 1}{-3x-1}\right)$$

$$\ =\ -3\left(\ \frac{\ -9x^2-3x-6x-2-1}{3x+1}\right)$$

$$\ =\ -3\left(\ \frac{\ -9\left(-3x-1\right)-9x-3}{3x+1}\right)$$

$$=-3\left(\ \frac{\ 27x+9-9x-3}{3x+1}\right)$$

$$=-3\left(\ \frac{\ 18x+6}{3x+1}\right)$$

$$=-3\times\ 6\left(\ \frac{\ 3x+1}{3x+1}\right)$$

$$=-18$$