If $$x + \frac{1}{x} = - 14,$$   and x < -1 what will be the value of $${x^2} - \frac{1}{{{x^2}}} = ?$$

$$\eqalign{ & x + \frac{1}{x} = - 14 \cr & x - \frac{1}{x} = \sqrt {{{\left( { - 14} \right)}^2} - 4} \cr & = \sqrt {196 - 4} \cr & = \sqrt {192} \cr & = 8\sqrt 3 \cr & \therefore \,x < - 1 \cr & \left( {x + \frac{1}{x}} \right)\left( {x - \frac{1}{x}} \right) = - 14x\left( {8\sqrt 3 } \right) \cr & {x^2} - \frac{1}{{{x^2}}} = + 112\sqrt 3 \cr} $$