Solving a particular function

By dividing numerator and denominator by $\omega_p^2$, and taking the $2$ inside the parentheses, you get $$ \frac{1}{(\sqrt{2}\omega/\omega_p-\sqrt{2}kv_0/\omega_p)^2}+\frac{1}{(\sqrt{2}\omega/\omega_p+\sqrt{2}kv_0/\omega_p)^2}=1 $$ The substitutions provide $$ \frac{1}{(y-\sqrt{2}\alpha)^2}+\frac{1}{(y+\sqrt{2}\alpha)^2}=1 $$ i.e. $$ \frac{(y+\sqrt{2}\alpha)^2+(y-\sqrt{2}\alpha)^2}{(y^2-2\alpha^2)^2}=1 $$ and again $$ \frac{2y^2+4\alpha^2}{(y^2-2\alpha^2)^2}=1 $$ Finally, solving the quadratic equation $$ (y^2-2\alpha^2)^2=2(y^2+2\alpha^2) $$ with respect to $y^2$ you get the result