Answer
$$\eqalign{
& {\bf{e}} = \sqrt 2 \cr
& {\bf{e}} \approx 1.4 \cr} $$
Work Step by Step
$$\eqalign{
& \frac{{{x^2}}}{8} - \frac{{{y^2}}}{8} = 1 \cr
& {\text{The equation is written in the form }}\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1 \cr
& \frac{{{x^2}}}{8} - \frac{{{y^2}}}{8} = 1,{\text{ then }}{a^2} = 8{\text{ and }}{b^2} = 8 \cr
& {\text{The excentricity of a hyperbola is giving by}} \cr
& {\bf{e}} = \frac{{\sqrt {{a^2} + {b^2}} }}{a} \cr
& {\bf{e}} = \frac{{\sqrt {8 + 8} }}{{\sqrt 8 }} \cr
& {\bf{e}} = \sqrt {\frac{{16}}{8}} \cr
& {\bf{e}} = \sqrt 2 \cr
& {\bf{e}} \approx 1.4 \cr} $$
